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FFIRS, , 12/15/2010, , 10:13:22, , Page 3, , INTRODUCTION TO REAL ANALYSIS, Fourth Edition, , Robert G. Bartle, Donald R. Sherbert, , University of Illinois, Urbana-Champaign, , John Wiley & Sons, Inc.

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FTOC, , 12/08/2010, , 15:45:54, , Page 11, , CONTENTS, , CHAPTER 1 PRELIMINARIES 1, 1.1, 1.2, 1.3, , Sets and Functions 1, Mathematical Induction 12, Finite and Infinite Sets 16, , CHAPTER 2 THE REAL NUMBERS 23, 2.1, 2.2, 2.3, 2.4, 2.5, , The Algebraic and Order Properties of R 23, Absolute Value and the Real Line 32, The Completeness Property of R 36, Applications of the Supremum Property 40, Intervals 46, , CHAPTER 3 SEQUENCES AND SERIES 54, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, , Sequences and Their Limits 55, Limit Theorems 63, Monotone Sequences 70, Subsequences and the Bolzano-Weierstrass Theorem 78, The Cauchy Criterion 85, Properly Divergent Sequences 91, Introduction to Infinite Series 94, , CHAPTER 4 LIMITS 102, 4.1, 4.2, 4.3, , Limits of Functions 103, Limit Theorems 111, Some Extensions of the Limit Concept 116, , CHAPTER 5 CONTINUOUS FUNCTIONS 124, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, , Continuous Functions 125, Combinations of Continuous Functions 130, Continuous Functions on Intervals 134, Uniform Continuity 141, Continuity and Gauges 149, Monotone and Inverse Functions 153, , xi

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FTOC, , 12/08/2010, , 15:45:54, , xii, , Page 12, , CONTENTS, , CHAPTER 6 DIFFERENTIATION 161, 6.1, 6.2, 6.3, 6.4, , The Derivative 162, The Mean Value Theorem 172, L’Hospital’s Rules 180, Taylor’s Theorem 188, , CHAPTER 7 THE RIEMANN INTEGRAL 198, 7.1, 7.2, 7.3, 7.4, 7.5, , Riemann Integral 199, Riemann Integrable Functions 208, The Fundamental Theorem 216, The Darboux Integral 225, Approximate Integration 233, , CHAPTER 8 SEQUENCES OF FUNCTIONS 241, 8.1, 8.2, 8.3, 8.4, , Pointwise and Uniform Convergence 241, Interchange of Limits 247, The Exponential and Logarithmic Functions 253, The Trigonometric Functions 260, , CHAPTER 9 INFINITE SERIES 267, 9.1, 9.2, 9.3, 9.4, , Absolute Convergence 267, Tests for Absolute Convergence 270, Tests for Nonabsolute Convergence 277, Series of Functions 281, , CHAPTER 10 THE GENERALIZED RIEMANN INTEGRAL 288, 10.1, 10.2, 10.3, 10.4, , Definition and Main Properties 289, Improper and Lebesgue Integrals 302, Infinite Intervals 308, Convergence Theorems 315, , CHAPTER 11 A GLIMPSE INTO TOPOLOGY 326, 11.1, 11.2, 11.3, 11.4, , Open and Closed Sets in R 326, Compact Sets 333, Continuous Functions 337, Metric Spaces 341, , APPENDIX A LOGIC AND PROOFS 348, APPENDIX B FINITE AND COUNTABLE SETS 357

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C04, , 12/09/2010, , 14:49:6, , Page 103, , 4.1 LIMITS OF FUNCTIONS, , 103, , Gottfried Leibniz, Gottfried Wilhelm Leibniz (1646–1716) was born in Leipzig, Germany., He was six years old when his father, a professor of philosophy, died and, left his son the key to his library and a life of books and learning. Leibniz, entered the University of Leipzig at age 15, graduated at age 17, and, received a Doctor of Law degree from the University of Altdorf four years, later. He wrote on legal matters, but was more interested in philosophy. He, also developed original theories about language and the nature of the, universe. In 1672, he went to Paris as a diplomat for four years. While, there he began to study mathematics with the Dutch mathematician Christiaan Huygens. His, travels to London to visit the Royal Academy further stimulated his interest in mathematics. His, background in philosophy led him to very original, though not always rigorous, results., Unaware of Newtons’s unpublished work, Leibniz published papers in the 1680s that, presented a method of finding areas that is known today as the Fundamental Theorem of Calculus., He coined the term ‘‘calculus’’ and invented the dy=dx and elongated S notations that are used, today. Unfortunately, some followers of Newton accused Leibniz of plagiarism, resulting in a, dispute that lasted until Leibniz’s death. Their approaches to calculus were quite different and it is, now evident that their discoveries were made independently. Leibniz is now renowned for his, work in philosophy, but his mathematical fame rests on his creation of the calculus., , Section 4.1 Limits of Functions, In this section we will introduce the important notion of the limit of a function. The, intuitive idea of the function f having a limit L at the point c is that the values f (x) are close, to L when x is close to (but different from) c. But it is necessary to have a technical way of, working with the idea of ‘‘close to’’ and this is accomplished in the e-d definition given, below., In order for the idea of the limit of a function f at a point c to be meaningful, it is, necessary that f be defined at points near c. It need not be defined at the point c, but it should, be defined at enough points close to c to make the study interesting. This is the reason for, the following definition., 4.1.1 Definition Let A
R. A point c 2 R is a cluster point of A if for every d > 0 there, exists at least one point x 2 A; x 6¼ c such that jx  cj < d., This definition is rephrased in the language of neighborhoods as follows: A point c is a, cluster point of the set A if every d-neighborhood V d ðcÞ ¼ ðc  d; c þ dÞ of c contains at, least one point of A distinct from c., Note The point c may or may not be a member of A, but even if it is in A, it is ignored, when deciding whether it is a cluster point of A or not, since we explicitly require that there, be points in V d ðcÞ \ A distinct from c in order for c to be a cluster point of A., For example, if A :¼ {1, 2}, then the point 1 is not a cluster point of A, since choosing, d :¼ 12 gives a neighborhood of 1 that contains no points of A distinct from 1. The same is, true for the point 2, so we see that A has no cluster points.

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C04, , 12/09/2010, , 14:49:6, , 104, , Page 104, , CHAPTER 4 LIMITS, , 4.1.2 Theorem A number c 2 R is a cluster point of a subset A of R if and only if there, exists a sequence (an) in A such that limðan Þ ¼ c and an 6¼ c for all n 2 N., Proof. If c is a cluster point of A, then for any n 2 N the (l=n)-neighborhood V 1=n ðcÞ, contains at least one point an in A distinct from c. Then an 2 A; an 6¼ c, and jan  cj < 1=n, implies limðan Þ ¼ c., Conversely, if there exists a sequence (an) in Anfcg with limðan Þ ¼ c, then, for any d > 0 there exists K such that if n  K, then an 2 V d ðcÞ. Therefore the, d-neighborhood V d ðcÞ of c contains the points an, for n  K, which belong to A and, are distinct from c., Q.E.D., The next examples emphasize that a cluster point of a set may or may not belong to, the set., 4.1.3 Examples (a) For the open interval A1 :¼ ð0; 1Þ, every point of the closed, interval [0,1] is a cluster point of A1. Note that the points 0, 1 are cluster points of A1,, but do not belong to A1. All the points of A1 are cluster points of A1., (b) A finite set has no cluster points., (c) The infinite set N has no cluster points., (d) The set A4 :¼ f1=n : n 2 N g has only the point 0 as a cluster point. None of the points, in A4 is a cluster point of A4., (e) If I :¼ ½0; 1, then the set A5 :¼ I \ Q consists of all the rational numbers in I. It follows, &, from the Density Theorem 2.4.8 that every point in I is a cluster point of A5., Having made this brief detour, we now return to the concept of the limit of a function at, a cluster point of its domain., The Definition of the Limit, We now state the precise definition of the limit of a function f at a point c. It is important to, note that in this definition, it is immaterial whether f is defined at c or not. In any case, we, exclude c from consideration in the determination of the limit., 4.1.4 Definition Let A
R, and let c be a cluster point of A. For a function f : A ! R, a, real number L is said to be a limit of f at c if, given any e > 0, there exists a d > 0 such that, if x 2 A and 0 < jx  cj < d, then j f ðxÞ  Lj < e., Remarks (a) Since the value of d usually depends on e, we will sometimes write dðeÞ, instead of d to emphasize this dependence., (b) The inequality 0 < jx  cj is equivalent to saying x 6¼ c., If L is a limit of f at c, then we also say that f converges to L at c. We often write, L ¼ lim f ðxÞ or, x!c, , L ¼ lim f :, x!c, , We also say that ‘‘f (x) approaches L as x approaches c.’’ (But it should be noted that the, points do not actually move anywhere.) The symbolism, f ðxÞ ! L, , as, , x!c, , is also used sometimes to express the fact that f has limit L at c.

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C04, , 12/09/2010, , 14:49:6, , Page 105, , 4.1 LIMITS OF FUNCTIONS, , 105, , If the limit of f at c does not exist, we say that f diverges at c., Our first result is that the value L of the limit is uniquely determined. This uniqueness, is not part of the definition of limit, but must be deduced., 4.1.5 Theorem If f : A ! R and if c is a cluster point of A, then f can have only one, limit at c., Proof. Suppose that numbers L and L0 satisfy Definition 4.1.4. For any e > 0, there exists, dðe=2Þ > 0 such that if x 2 A and 0 < jx  cj < dðe=2Þ, then j f ðxÞ  Lj < e=2. Also there, exists d0 ðe=2Þ such that if x 2 A and 0 < jx  cj < d0 ðe=2Þ, then j f ðxÞ  L0 j < e=2. Now, let d :¼ inf fdðe=2Þ; d0 ðe=2Þg. Then if x 2 A and 0 < jx  cj < d, the Triangle Inequality, implies that,
,
, jL  L0 j 
L  f ðxÞ
þ j f ðxÞ  L0 j < e=2 þ e=2 ¼ e:, Since e > 0 is arbitrary, we conclude that L  L0 ¼ 0, so that L ¼ L0 ., , Q.E.D., , The definition of limit can be very nicely described in terms of neighborhoods. (See, Figure 4.1.1.) We observe that because, V d ðcÞ ¼ ðc  d; c þ dÞ ¼ fx : jx  cj < dg;, the inequality 0 < jx  cj < d is equivalent to saying that x 6¼ c and x belongs to the, d-neighborhood V d ðcÞ of c. Similarly, the inequality j f ðxÞ  Lj < e is equivalent to saying, that f (x) belongs to the e-neighborhood V e ðLÞ of L. In this way, we obtain the following, result. The reader should write out a detailed argument to establish the theorem., , Figure 4.1.1, , The limit of f at c is L, , 4.1.6 Theorem Let f : A ! R and let c be a cluster point of A. Then the following, statements are equivalent., (i) lim f ðxÞ ¼ L., x!c, (ii) Given any e-neighborhood V e ðLÞ of L, there exists a d-neighborhood V d ðcÞ of c such, that if x 6¼ c is any point in V d ðcÞ \ A, then f (x) belongs to V e ðLÞ., We now give some examples that illustrate how the definition of limit is applied., 4.1.7 Examples (a) lim b ¼ b., x!c, , To be more explicit, let f ðxÞ :¼ b for all x 2 R. We want to show that lim f ðxÞ ¼ b. If, x!c, e > 0 is given, we let d :¼ 1. (In fact, any strictly positive d will serve the purpose.) Then if

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C04, , 12/09/2010, , 14:49:7, , Page 107, , 4.1 LIMITS OF FUNCTIONS, , 107, , In order to make this last term less than e it suffices to take jx  cj < 12 c2 e. Consequently, if, we choose, , , 1 1 2, dðeÞ :¼ inf c; c e ;, 2 2, 1, then if 0 < jx  cj < dðeÞ, it will,  follow first that jx  cj < 2 c so that (2) is valid, and, 1 2, therefore, since jx  cj < 2 c e, that,
,

,
,
,

,
,
wðxÞ  1
¼
1  1
< e:,
,
,
, c, x c
, , Since we have a way of choosing dðeÞ > 0 for an arbitrary choice of e > 0, we infer that, lim w ¼ 1=c., , x!c, , x3  4 4, ¼ :, x!2 x2 þ 1, 5, Let cðxÞ :¼ ðx3  4Þ=ðx2 þ 1Þ for x 2 R. Then a little algebraic manipulation, gives us,
,
,
, 4


5x3  4x2  24
,
,
cðxÞ 
¼, 5, 5 ð x2 þ 1 Þ,
3,
,
5x þ 6x þ 12
,  jx  2j:, ¼, 5 ð x2 þ 1 Þ, , (e) lim, , To get a bound on the coefficient of jx  2j, we restrict x by the condition 1 < x < 3., For x in this interval, we have 5x2 þ 6x þ 12  5  32 þ 6  3 þ 12 ¼ 75 and, 5ðx2 þ 1Þ  5ð1 þ 1Þ ¼ 10, so that,
,
,
,
,
cðxÞ  4
 75 jx  2j ¼ 15 jx  2j:,
, 5
10, 2, Now for given e > 0, we choose, , , , 2, e :, dðeÞ :¼ inf 1;, 15, , Then if 0 < jx  2j < dðeÞ, we have jcðxÞ  ð4=5Þj  ð15=2Þjx  2j < e. Since e > 0 is, &, arbitrary, the assertion is proved., Sequential Criterion for Limits, The following important formulation of limit of a function is in terms of limits of, sequences. This characterization permits the theory of Chapter 3 to be applied to the, study of limits of functions., 4.1.8 Theorem (Sequential Criterion) Let f : A ! R and let c be a cluster point of A., Then the following are equivalent., (i) lim f ¼ L:, x!c, (ii) For every sequence (xn) in A that converges to c such that xn 6¼ c for all n 2 N, the, sequence ( f(xn)) converges to L., Proof. (i) ) (ii). Assume f has limit L at c, and suppose (xn) is a sequence in A with, limðxn Þ ¼ c and xn 6¼ c for all n. We must prove that the sequence ð f ðxn ÞÞ converges to L., Let e > 0 be given. Then by Definition 4.1.4, there exists d > 0 such that if x 2 A satisfies

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C04, , 12/09/2010, , 14:49:8, , 108, , Page 108, , CHAPTER 4 LIMITS, , 0 < jx  cj < d, then f (x) satisfies j f ðxÞ  Lj < e. We now apply the definition of, convergent sequence for the given d to obtain a natural number K(d) such that if n >, K ðdÞ then jxn  cj < d. But for each such xn we have j f ðxn Þ  Lj < e. Thus if n > K ðdÞ,, then j f ðxn Þ  Lj < e. Therefore, the sequence ð f ðxn ÞÞ converges to L., (ii) ) (i). [The proof is a contrapositive argument.] If (i) is not true, then there exists, an e0 -neighborhood V e0 ðLÞ such that no matter what d-neighborhood of c we pick, there, will be at least one number xd in A \ V d ðcÞ with xd 6¼ c such that f ðxd Þ 2, = V e0 ðLÞ. Hence for, every n 2 N, the (1=n)-neighborhood of c contains a number xn such that, 0 < jxn  cj < 1=n, , and, , xn 2 A ;, , for all, , n2N:, , but such that, j f ð xn Þ  L j  e 0, , We conclude that the sequence (xn) in An{c} converges to c, but the sequence ð f ðxn ÞÞ does, not converge to L. Therefore we have shown that if (i) is not true, then (ii) is not true. We, Q.E.D., conclude that (ii) implies (i)., We shall see in the next section that many of the basic limit properties of functions can, be established by using corresponding properties for convergent sequences. For example,, we know from our, with sequences that if (xn) is any sequence that converges to a,  work, , number c, then x2n converges to c2. Therefore, by the sequential criterion, we can, conclude that the function hðxÞ :¼ x2 has limit lim hðxÞ ¼ c2 ., x!c, , Divergence Criteria, It is often important to be able to show (i) that a certain number is not the limit of a function, at a point, or (ii) that the function does not have a limit at a point. The following result is a, consequence of (the proof of) Theorem 4.1.8. We leave the details of its proof as an, important exercise., 4.1.9 Divergence Criteria Let A
R, let f : A ! R and let c 2 R be a cluster, point of A., (a) If L 2 R, then f does not have limit L at c if and only if there exists a sequence (xn) in A, with xn 6¼ c for all n 2 N such that the sequence (xn) converges to c but the sequence, ð f ðxn ÞÞ does not converge to L., (b) The function f does not have a limit at c if and only if there exists a sequence (xn) in A, with xn 6¼ c for all n 2 N such that the sequence (xn) converges to c but the sequence, ð f ðxn ÞÞ does not converge in R., We now give some applications of this result to show how it can be used., 4.1.10 Examples (a) lim ð1=xÞ does not exist in R., x!0, , As in Example 4.1.7(d), let wðxÞ :¼ 1=x for x > 0. However, here we consider c ¼ 0., The argument given in Example 4.1.7(d) breaks down if c ¼ 0 since we cannot obtain, a bound such as that in (2) of that example. Indeed, if we take the sequence (xn) with, xn :¼ 1=n for n 2 N, then limðxn Þ ¼ 0, but wðxn Þ ¼ 1=ð1=nÞ ¼ n. As we know, the, sequence ðwðxn ÞÞ ¼ ðnÞ is not convergent in R, since it is not bounded. Hence, by, Theorem 4.1.9(b), lim ð1=xÞ does not exist in R., x!0, (b) lim sgnðxÞ does not exist., x!0

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C04, , 12/09/2010, , 14:49:9, , Page 109, , 4.1 LIMITS OF FUNCTIONS, , Let the signum function sgn be defined by, 8, for, < þ1, sgnðxÞ :¼, 0, for, :, 1, for, , 109, , x > 0;, x ¼ 0;, x < 0:, , Note that sgnðxÞ ¼ x=jxj for x 6¼ 0. (See Figure 4.1.2.) We shall show that sgn does not, have a limit at x ¼ 0. We shall do this by showing that there is a sequence (xn) such that, limðxn Þ ¼ 0, but such that ðsgnðxn ÞÞ does not converge., , Figure 4.1.2, , The signum function, , Indeed, let xn :¼ ð1Þn =n for n 2 N so that limðxn Þ ¼ 0. However, since, sgnðxn Þ ¼ ð1Þn, , for, , n2N;, , it follows from Example 3.4.6(a) that ðsgnðxn ÞÞ does not converge. Therefore lim sgnðxÞ, x!0, does not exist., (c)y lim sinð1=xÞ does not exist in R., x!0, , Let gðxÞ :¼ sinð1=xÞ for x 6¼ 0. (See Figure 4.1.3.) We shall show that g does not have, a limit at c ¼ 0, by exhibiting two sequences (xn) and (yn) with xn 6¼ 0 and yn 6¼ 0 for all, n 2 N and such that limðxn Þ ¼ 0 and limðyn Þ ¼ 0, but such that limðgðxn ÞÞ 6¼ limðgðyn ÞÞ., In view of Theorem 4.1.9 this implies that lim g cannot exist. (Explain why.), x!0, , Figure 4.1.3, , The function gðxÞ ¼ sinð1=xÞðx 6¼ 0Þ, , Indeed, we recall from calculus that sin t ¼ 0 if t ¼ np for n 2 Z, and that sin t ¼ þ1, if t ¼ 12 p þ 2p n for n 2 Z. Now let xn :¼ 1=np for n 2 N; then limðxn Þ ¼ 0 and gðxn Þ ¼, np ¼ 0  for all n 2 N, so that limðgðxn ÞÞ ¼ 0. On the other hand,, sin,  let yn :¼, 1, 1, 1, p, þ, 2pn, for, n, 2, N;, then, lim, ð, y, Þ, ¼, 0, and, g, ð, y, Þ, ¼, sin, p, þ, 2p, n, ¼ 1 for all, n, n, 2, 2, &, n 2 N, so that limðgðyn ÞÞ ¼ 1. We conclude that lim sinð1=xÞ does not exist., x!0, , y, , In order to have some interesting applications in this and later examples, we shall make use of well-known, properties of trigonometric and exponential functions that will be established in Chapter 8.

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C04, , 12/09/2010, , 14:49:10, , 110, , Page 110, , CHAPTER 4 LIMITS, , Exercises for Section 4.1, 1. Determine, a
condition on jx  1j that will assure that:
,
,
, (a)
x2  1
< 12,, (b)
x2  1
< 1=103 ,,
,
,
,
, for a given n 2 N ,, (d)
x3  1
< 1=n, for a given n 2 N ., (c)
x2  1
< 1=n, 2. Determine a condition on jx  4j that will assure that:, pffiffiffi, pffiffiffi, (a) j x  2j < 12,, (b) j x  2j < 102 ., 3. Let c be a cluster point of A
R and let f : A ! R. Prove that lim f ðxÞ ¼ L if and only if, x!c, lim j f ðxÞ  Lj ¼ 0., x!c, , 4. Let f :¼ R ! R and let c 2 R. Show that lim f ðxÞ ¼ L if and only if lim f ðx þ cÞ ¼ L., x!c, , x!0, , 2, 5. Let, points x, c 2 I, show that,
I :¼ ð20;,
aÞ where a > 0, and let gðxÞ :¼ x for x 2 I. For any,
gðxÞ  c
 2ajx  cj. Use this inequality to prove that lim x2 ¼ c2 for any c 2 I., x!c, , 6. Let I be an interval in R, let f : I ! R, and let c 2 I. Suppose there exist constants K and L such, that j f ðxÞ  Lj  K jx  cj for x 2 I. Show that lim f ðxÞ ¼ L., x!c, , 7. Show that lim x3 ¼ c3 for any c 2 R., x!c pffiffiffi, pffiffiffi, 8. Show that lim x ¼ c for any c > 0., x!c, , 9. Use either the e-d definition of limit or the Sequential Criterion for limits, to establish the, following limits., 1, x, 1, ¼ 1,, (b) lim, ¼ ,, (a) lim, x!2 1  x, x!1 1 þ x, 2, x2, x2  x þ 1 1, ¼ ., ¼ 0,, (d) lim, (c) lim, x!0 jxj, x!1, xþ1, 2, 10. Use the definition of limit to show that, , , (a) lim x2 þ 4x ¼ 12,, , (b), , x!2, , 11. Use the definition of limit to prove the following., 2x þ 3, (a) lim, ¼ 3,, (b), x!3 4x  9, 12. Show that the following limits do not exist., 1, (b), (a) lim 2 ðx > 0Þ ,, x!0 x, (d), (c) lim ðx þ sgnðxÞÞ,, x!0, , lim, , xþ5, ¼ 4., þ3, , x!1 2x, , lim, , x!6, , x2  3x, ¼ 2:, xþ3, , 1, lim pffiffiffi ðx > 0Þ ,, x, , lim sin 1=x2 ., x!0, , x!0, , 13. Suppose the function f : R ! R has limit L at 0, and let a > 0. If g : R ! R is defined by, gðxÞ :¼ f ðaxÞ for x 2 R, show that lim gðxÞ ¼ L., x!0, , 14. Let c 2 R and let f : R ! R be such that lim ð f ðxÞÞ2 ¼ L., x!c, , (a) Show that if L ¼ 0, then lim f ðxÞ ¼ 0., x!c, , (b) Show by example that if L 6¼ 0, then f may not have a limit at c., 15. Let f : R ! R be defined by setting f ðxÞ :¼ x if x is rational, and f ðxÞ ¼ 0 if x is irrational., (a) Show that f has a limit at x ¼ 0., (b) Use a sequential argument to show that if c 6¼ 0, then f does not have a limit at c., 16. Let f : R ! R, let I be an open interval in R, and let c 2 I. If f1 is the restriction of f to I, show, that f1 has a limit at c if and only if f has a limit at c, and that the limits are equal., 17. Let f : R ! R, let J be a closed interval in R, and let c 2 J. If f2 is the restriction of f to J,, show that if f has a limit at c then f2 has a limit at c. Show by example that it does not follow that, if f2 has a limit at c, then f has a limit at c.

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C04, , 12/09/2010, , 14:49:10, , Page 111, , 4.2 LIMIT THEOREMS, , 111, , Section 4.2 Limit Theorems, We shall now obtain results that are useful in calculating limits of functions. These results, are parallel to the limit theorems established in Section 3.2 for sequences. In fact, in most, cases these results can be proved by using Theorem 4.1.8 and results from Section 3.2., Alternatively, the results in this section can be proved by using e-d arguments that are very, similar to the ones employed in Section 3.2., 4.2.1 Definition Let A
R, let f : A ! R, and let c 2 R be a cluster point of A. We say, that f is bounded on a neighborhood of c if there exists a d-neighborhood V d ðcÞ of c and a, constant M > 0 such that we have j f ðxÞj  M for all x 2 A \ V d ðcÞ., 4.2.2 Theorem If A
R and f : A ! R has a limit at c 2 R, then f is bounded on some, neighborhood of c., Proof. If L :¼ lim f , then for e ¼ 1, there exists d > 0 such that if 0 < jx  cj < d, then, x!c, , j f ðxÞ  Lj < 1; hence (by Corollary 2.2.4(a)),, j f ðxÞj  jLj  j f ðxÞ  Lj < 1:, Therefore, if x 2 A \ V d ðcÞ; x 6¼ c, then j f ðxÞj  jLj þ 1. If c 2, = A, we take M ¼ jLj þ 1,, while if c 2 A we take M :¼ supfj f ðcÞj; jLj þ 1g. It follows that if x 2 A \ V d ðcÞ, then, j f ðxÞj  M. This shows that f is bounded on the neighborhood V d ðcÞ of c., Q.E.D., The next definition is similar to the definition for sums, differences, products, and, quotients of sequences given in Section 3.2., 4.2.3 Definition Let A
R and let f and g be functions defined on A to R. We define the, sum f þ g, the difference f  g, and the product fg on A to R to be the functions given by, ð f þ gÞðxÞ :¼ f ðxÞ þ gðxÞ;, ð f  gÞðxÞ :¼ f ðxÞ  gðxÞ;, ð f gÞðxÞ :¼ f ðxÞgðxÞ, for all x 2 A. Further, if b 2 R, we define the multiple bf to be the function given by, ðbf ÞðxÞ :¼ bf ðxÞ, , for all x 2 A :, , Finally, if hðxÞ 6¼ 0 for x 2 A, we define the quotient f=h to be the function given by, , f, f ðxÞ, for all, x 2 A:, ðxÞ :¼, h, hðxÞ, 4.2.4 Theorem Let A
R, let f and g be functions on A to R, and let c 2 R be a cluster, point of A. Further, let b 2 R., (a) If lim f ¼ L and lim g ¼ M, then:, x!c, , x!c, , lim ð f þ gÞ ¼ L þ M;, , x!c, , lim ð f gÞ ¼ LM;, , x!c, , lim ð f  gÞ ¼ L  M;, , x!c, , lim ðbf Þ ¼ bL:, , x!c, , (b) If h : A ! R, if hðxÞ 6¼ 0 for all x 2 A, and if lim h ¼ H 6¼ 0, then, x!c, , f, L, ¼ :, lim, x!c h, H

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C04, , 12/09/2010, , 14:49:11, , 114, , Page 114, , CHAPTER 4 LIMITS, , If c is not a zero of q(x), then qðcÞ 6¼ 0, and it follows from part (f) that lim qðxÞ ¼ qðcÞ 6¼ 0., x!c, Therefore we can apply Theorem 4.2.4(b) to conclude that, lim pðxÞ pðcÞ, pðxÞ x!c, ¼, :, ¼, x!c qðxÞ, lim qðxÞ qðcÞ, , &, , lim, , x!c, , The next result is a direct analogue of Theorem 3.2.6., 4.2.6 Theorem Let A
R, let f : A ! R, and let c 2 R be a cluster point of A. If, a  f ðxÞ  b, , f or all, , x 2 A; x 6¼ c;, , and if lim f exists, then a  lim f  b., x!c, , x!c, , Proof. Indeed, if L ¼ lim f , then it follows from Theorem 4.1.8 that if (xn) is any, x!c, sequence of real numbers such that c 6¼ xn 2 A for all n 2 N and if the sequence (xn), converges to c, then the sequence ð f ðxn ÞÞ converges to L. Since a  f ðxn Þ  b for all, n 2 N, it follows from Theorem 3.2.6 that a  L  b., Q.E.D., We now state an analogue of the Squeeze Theorem 3.2.7. We leave its proof to the reader., 4.2.7 Squeeze Theorem Let A
R, let f, g, h: A ! R, and let c 2 R be a cluster point, of A. If, f ðxÞ  gðxÞ  hðxÞ f or all, , x 2 A; x 6¼ c;, , and if lim f ¼ L ¼ lim h, then lim g ¼ L., x!c, , x!c, , x!c, , 4.2.8 Examples (a) lim x3=2 ¼ 0 ðx > 0Þ., x!0, , Let f ðxÞ :¼ x3=2 for x > 0. Since the inequality x < x1=2  1 holds for 0 < x  1, (why?), it follows that x2  f ðxÞ ¼ x3=2  x for 0 < x  1. Since, lim x2 ¼ 0 and, , x!0, , lim x ¼ 0;, , x!0, , it follows from the Squeeze Theorem 4.2.7 that lim x3=2 ¼ 0., x!0, , (b) lim sin x ¼ 0., x!0, , It will be proved later (see Theorem 8.4.8), that, x  sin x  x, , for all, , x  0:, , Since lim ðxÞ ¼ 0, it follows from the Squeeze Theorem that lim sin x ¼ 0., x!0, , x!0, , (c) lim cos x ¼ 1., x!0, It will be proved later (see Theorem 8.4.8) that, ð1Þ, , 1  12 x2  cos x  1 for all, , x 2 R:, , Since lim 1  12 x2 ¼ 1, it follows from the Squeeze Theorem that lim cos x ¼ 1., x!0, x!0, , cos x  1, (d) lim, ¼ 0., x!0, x, 

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C04, , 12/09/2010, , 14:49:12, , Page 115, , 4.2 LIMIT THEOREMS, , 115, , We cannot use Theorem 4.2.4(b) to evaluate this limit. (Why not?) However, it follows, from the inequality (1) in part (c) that,  12 x  ðcos x  1Þ=x  0, , for, , x>0, , 0  ðcos x  1Þ=x   12 x, , for, , x < 0:, , and that, Now let f ðxÞ :¼ x=2 for x  0 and f ðxÞ :¼ 0 for x < 0, and let hðxÞ :¼ 0 for x  0 and, hðxÞ :¼ x=2 for x < 0. Then we have, f ðxÞ  ðcos x  1Þ=x  hðxÞ, , for, , x 6¼ 0:, , Since it is readily seen that lim f ¼ 0 ¼ lim h, it follows from the Squeeze Theorem that, x!0, x!0, lim ðcos x  1Þ=x ¼ 0., , x!0, sin x, (e) lim, ¼ 1., x!0, x, Again we cannot use Theorem 4.2.4(b) to evaluate this limit. However, it will be, proved later (see Theorem 8.4.8) that, x  16 x3  sin x  x for x  0, and that, x  sin x  x  16 x3, , for, , x  0:, , Therefore it follows (why?) that, 1  16 x2  ðsin xÞ=x  1 for all x 6¼ 0:, , , But since lim 1  16 x2 ¼ 1  16  lim x2 ¼ 1, we infer from the Squeeze Theorem that, x!0, , x!0, , lim ðsin xÞ=x ¼ 1., , x!0, , (f) lim ðx sinð1=xÞÞ ¼ 0., x!0, , Let f ðxÞ ¼ x sinð1=xÞ for x ¼, 6 0. Since 1  sin z  1 for all z 2 R, we have the, inequality, jxj  f ðxÞ ¼ x sinð1=xÞ  jxj, for all x 2 R, x 6¼ 0. Since lim jxj ¼ 0, it follows from the Squeeze Theorem that lim f ¼ 0., x!0, , x!0, , For a graph, see Figure 5.1.3 or the cover of this book., , &, , There are results that are parallel to Theorems 3.2.9 and 3.2.10; however, we will leave, them as exercises. We conclude this section with a result that is, in some sense, a partial, converse to Theorem 4.2.6., 4.2.9 Theorem Let A
R, let f : A ! R and let c 2 R be a cluster point of A. If, h, i, lim f > 0 respectively; lim f < 0 ;, x!c, , x!c, , then there exists a neighborhood Vd(c) of c such that f (x) > 0 [respectively, f (x) < 0] for, all x 2 A \ V d ðcÞ, x 6¼ c., Proof. Let L :¼ lim f and suppose that L > 0. We take e ¼ 12 L > 0 in Definition 4.1.4, and, x!c, , obtain a number d > 0 such that if 0 < jx  cj < d and x 2 A, then j f ðxÞ  Lj < 12 L., Therefore (why?) it follows that if x 2 A \ V d ðcÞ, x 6¼ c, then f ðxÞ > 12 L > 0., If L < 0, a similar argument applies., Q.E.D.

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C04, , 12/09/2010, , 14:49:13, , 116, , Page 116, , CHAPTER 4 LIMITS, , Exercises for Section 4.2, 1. Apply Theorem 4.2.4 to determine the following limits:, x2 þ 2, lim ðx þ 1Þð2x þ 3Þ ðx 2 R Þ;, (a) x!1, (b) lim, ðx > 0Þ;, x!1 x2  2, , 1, 1, xþ1, , ðx > 0Þ ;, (d) lim 2, (c) lim, ðx 2 R Þ:, x!2 x þ 1, x!0 x þ 2, 2x, 2. Determine the following limits and state which theorems are used in each case. (You may wish, to use Exercise 15 below.), rffiffiffiffiffiffiffiffiffiffiffiffiffiffi, x2  4, 2x þ 1, (b) lim, (a) lim, ðx > 0Þ;, ðx > 0Þ;, x!2 x  2, x!2, xþ3, pffiffiffi, 2, x1, (d) lim, (c) lim ðx þ 1Þ  1 ðx > 0Þ;, ðx > 0Þ:, x!1 x  1, x!0, x, pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi, 1 þ 2x  1 þ 3x, 3. Find lim, where x > 0., x!0, x þ 2x2, 4. Prove that lim cosð1=xÞ does not exist but that lim x cosð1=xÞ ¼ 0., x!0, , x!0, , 5. Let f, g be defined on A
R to R, and let c be a cluster point of A. Suppose that f is bounded on a, neighborhood of c and that lim g ¼ 0. Prove that lim f g ¼ 0., x!c, x!c, 6. Use the definition of the limit to prove the first assertion in Theorem 4.2.4(a)., 7. Use the sequential formulation of the limit to prove Theorem 4.2.4(b)., 8. Let n 2 N be such that n  3. Derive the inequality x2  xn  x2 for 1 < x < 1. Then use, the fact that lim x2 ¼ 0 to show that lim xn ¼ 0., x!0, , x!0, , 9. Let f, g be defined on A to R and let c be a cluster point of A., (a) Show that if both lim f and lim ð f þ gÞ exist, then lim g exists., x!c, , x!c, , x!c, , (b) If lim f and lim f g exist, does it follow that lim g exists?, x!c, , x!c, , x!c, , 10. Give examples of functions f and g such that f and g do not have limits at a point c, but such that, both f þ g and fg have limits at c., 11. Determine whether the following limits exist in R., , , , , lim sin 1=x2, ðx 6¼ 0Þ,, lim x sin 1=x2, ðx 6¼ 0Þ,, (a) x!0, (b) x!0, , pffiffiffi , 2, lim sgn sinð1=xÞ ðx 6¼ 0Þ,, ðx > 0Þ., lim x sin 1=x, (c) x!0, (d) x!0, 12. Let f : R ! R be such that f ðx þ yÞ ¼ f ðxÞ þ f ðyÞ for all x, y in R. Assume that lim f ¼ L, x!0, , exists. Prove that L ¼ 0, and then prove that f has a limit at every point c 2 R. [Hint: First note that, f ð2xÞ ¼ f ðxÞ þ f ðxÞ ¼ 2f ðxÞ for x 2 R. Also note that f ðxÞ ¼ f ðx  cÞ þ f ðcÞ for x, c in R.], 13. Functions f and g are defined on R by f (x) :¼ x þ 1 and g (x) :¼ 2 if x 6¼ 1 and g(1) :¼ 0., (a) Find lim g (f (x)) and compare with the value of g (lim f (x))., x!1, , x!1, , (b) Find lim f (g (x)) and compare with the value of f (lim g (x))., x!1, , x!1, , 14. Let A
R, let f : A ! R and let c 2 R be a cluster point of A. If lim f exists, and if j f j denotes,
,
, x!c,
,
, the function defined for x 2 A by jf jðxÞ :¼ j f ðxÞj, prove that lim jf j ¼
lim f
., x!c, , x!c, , 15. Let A
R, let f : A ! R, andplet, suppose, pffiffiffiffiffiffiffiffithat, ffi, ffiffiffi c 2 R be a cluster point of A. In addition,, pffiffiffi, f ðxÞ  0 for all x 2 A, and p, letffiffiffi f q, beffiffiffiffiffiffiffiffiffiffi, the function defined for x 2 A by ð f Þ (x) :¼ f ðxÞ. If, lim f exists, prove that lim f ¼ lim f ., x!c, , x!c, , x!c, , Section 4.3 Some Extensions of the Limit Concepty, In this section, we shall present three types of extensions of the notion of a limit of a, function that often occur. Since all the ideas here are closely parallel to ones we have, already encountered, this section can be read easily., y, , This section can be largely omitted on a first reading of this chapter.

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C04, , 12/09/2010, , 14:49:13, , Page 117, , 4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT, , 117, , One-Sided Limits, There are times when a function f may not possess a limit at a point c, yet a limit, does exist when the function is restricted to an interval on one side of the cluster, point c., For example, the signum function considered in Example 4.1.10(b), and illustrated in Figure 4.1.2, has no limit at c ¼ 0. However, if we restrict the signum function, to the interval (0, 1), the resulting function has a limit of 1 at c ¼ 0. Similarly, if, we restrict the signum function to the interval (1, 0), the resulting function has a limit, of 1 at c ¼ 0. These are elementary examples of right-hand and left-hand limits at, c ¼ 0., 4.3.1 Definition Let A 2 R and let f : A ! R., (i), , If c 2 R is a cluster point of the set A \ ðc; 1Þ ¼ fx 2 A: x > cg, then we say that, L 2 R is a right-hand limit of f at c and we write, lim f ¼ L, , x!cþ, , or, , lim f ðxÞ ¼ L, , x!cþ, , if given any e > 0 there exists a d ¼ dðeÞ > 0 such that for all x 2 A with, 0 < x  c < d, then j f ðxÞ  Lj < e., (ii) If c 2 R is a cluster point of the set A \ ð1; cÞ ¼ fx 2 A: x < cg, then we say that, L 2 R is a left-hand limit of f at c and we write, lim f ¼ L, , x!c, , or, , lim f ðxÞ ¼ L, , x!c, , if given any e > 0 there exists a d > 0 such that for all x 2 A with 0 < c  x < d,, then j f ðxÞ  Lj < e., Notes (1) The limits lim f and lim f are called one-sided limits of f at c. It is possible, x!cþ, x!c, that neither one-sided limit may exist. Also, one of them may exist without the other, existing. Similarly, as is the case for f ðxÞ :¼ sgnðxÞ at c ¼ 0, they may both exist and be, different., (2) If A is an interval with left endpoint c, then it is readily seen that f : A ! R has a limit, at c if and only if it has a right-hand limit at c. Moreover, in this case the limit lim f and the, x!c, right-hand limit lim f are equal. (A similar situation occurs for the left-hand limit when A, x!cþ, is an interval with right endpoint c.), The reader can show that f can have only one right-hand (respectively, left-hand) limit, at a point. There are results analogous to those established in Sections 4.1 and 4.2 for twosided limits. In particular, the existence of one-sided limits can be reduced to sequential, considerations., 4.3.2 Theorem Let A
R, let f : A ! R, and let c 2 R be a cluster point of A \ ðc; 1Þ., Then the following statements are equivalent:, (i), (ii), , lim f ¼ L., , x!cþ, , For every sequence (xn) that converges to c such that xn 2 A and xn > c for all, n 2 N, the sequence ð f ðxn ÞÞ converges to L.

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C04, , 12/09/2010, , 14:49:13, , 118, , Page 118, , CHAPTER 4 LIMITS, , Figure4.3.2 Graph of, hðxÞ ¼ 1= e1=x þ 1 ðx 6¼ 0Þ, , Figure 4.3.1 Graph of, gðxÞ ¼ e1=x ðx 6¼ 0Þ, , We leave the proof of this result (and the formulation and proof of the analogous, result for left-hand limits) to the reader. We will not take the space to write out the, formulations of the one-sided version of the other results in Sections 4.1 and 4.2., The following result relates the notion of the limit of a function to one-sided limits. We, leave its proof as an exercise., 4.3.3 Theorem Let A
R, let f : A ! R, and let c 2 R be a cluster point of both, of the sets A \ ðc; 1Þ and A \ ð1; cÞ. Then lim f ¼ L if and only if, x!c, lim f ¼ L ¼ lim f ., , x!cþ, , x!c, , 4.3.4 Examples (a) Let f ðxÞ :¼ sgnðxÞ., We have seen in Example 4.1.10(b) that sgn does not have a limit at 0. It is clear that, lim sgnðxÞ ¼ þ1 and that lim sgnðxÞ ¼ 1. Since these one-sided limits are different, it, , x!0þ, , x!0, , also follows from Theorem 4.3.3 that sgn(x) does not have a limit at 0., (b) Let gðxÞ :¼ e1=x for x 6¼ 0. (See Figure 4.3.1.), We first show that g does not have a finite right-hand limit at c ¼ 0 since it is, not bounded on any right-hand neighborhood ð0; dÞ of 0. We shall make use of the, inequality, 0 < t < et, , ð1Þ, , for t > 0;, , which will be proved later (see Corollary 8.3.3). It follows from (1) that if x > 0, then, 0 < 1=x < e1=x . Hence, if we take xn ¼ 1=n, then gðxn Þ > n for all n 2 N. Therefore, lim e1=x does not exist in R., x!0þ, However, lim e1=x ¼ 0. Indeed, if x < 0 and we take t ¼ 1=x in (1) we obtain, x!0, , 0 < 1=x < e1=x . Since x < 0, this implies that 0 < e1=x < x for all x < 0. It follows, from this inequality that lim e1=x ¼ 0., x!0, , , (c) Let hðxÞ :¼ 1= e1=x þ 1 for x 6¼ 0. (See Figure 4.3.2.), We have seen in part (b) that 0 < 1=x < e1=x for x > 0, whence, 0<, which implies that lim h ¼ 0., x!0þ, , 1, 1, < 1=x < x;, þ1 e, , e1=x

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C04, , 12/09/2010, , 14:49:14, , Page 119, , 4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT, , 119, , Since we have seen in part (b) that lim e1=x ¼ 0, it follows from the analogue of, x!0, Theorem 4.2.4(b) for left-hand limits that, , 1, 1, 1, lim, ¼ 1:, ¼, ¼, x!0 e1=x þ 1, lim e1=x þ 1 0 þ 1, x!0, , Note that for this function, both one-sided limits exist in R, but they are unequal., , &, , Infinite Limits, The function f ðxÞ :¼ 1=x2 for x 6¼ 0 (see Figure 4.3.3) is not bounded on a neighborhood, of 0, so it cannot have a limit in the sense of Definition 4.1.4. While the symbols, 1ð¼ þ1Þ and 1 do not represent real numbers, it is sometimes useful to be able to, say that ‘‘f ðxÞ ¼ 1=x2 tends to 1 as x ! 0.’’ This use of 1 will not cause any, difficulties, provided we exercise caution and never interpret 1 or 1 as being real, numbers., , Figure 4.3.3, f ðxÞ ¼ 1=x2, , Graph of, ðx 6¼ 0Þ, , Figure 4.3.4 Graph of, gðxÞ ¼ 1=x ðx 6¼ 0Þ, , 4.3.5 Definition Let A
R, let f : A ! R, and let c 2 R be a cluster point of A., (i), , We say that f tends to 1 as x ! c, and write, lim f ¼ 1;, , x!c, , (ii), , if for every a 2 R there exists d ¼ dðaÞ > 0 such that for all x 2 A with, 0 < jx  cj < d, then f ðxÞ > a., We say that f tends to 1 as x ! c, and write, lim f ¼ 1;, , x!c, , if for every b 2 R there exists d ¼ dðbÞ > 0 such that for all x 2 A with, 0 < jx  cj < d, then f ðxÞ < b., , , 4.3.6 Examples (a) lim 1=x2 ¼ 1., x!0, pffiffiffi, For, if a > 0 is given, let d :¼ 1= a. It follows that if 0 < jxj < d, then x2 < 1=a so, that 1=x2 > a., (b) Let gðxÞ :¼ 1=x for x 6¼ 0. (See Figure 4.3.4.)

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C04, , 12/09/2010, , 14:49:15, , 120, , Page 120, , CHAPTER 4 LIMITS, , The function g does not tend to either 1 or 1 as x ! 0. For, if a > 0 then gðxÞ < a, for all x < 0, so that g does not tend to 1 as x ! 0. Similarly, if b < 0 then gðxÞ > b for all, &, x > 0, so that g does not tend to 1 as x ! 0., While many of the results in Sections 4.1 and 4.2 have extensions to this limiting, notion, not all of them do since 1 are not real numbers. The following result is an, analogue of the Squeeze Theorem 4.2.7. (See also Theorem 3.6.4.), 4.3.7 Theorem Let A
R, let f ; g : A ! R, and let c 2 R be a cluster point of A., Suppose that f ðxÞ  gðxÞ for all x 2 A; x 6¼ c., (a) If lim f ¼ 1, then lim g ¼ 1., x!c, , x!c, , (b) If lim g ¼ 1, then lim f ¼ 1., x!c, , x!c, , Proof. (a) If lim f ¼ 1 and a 2 R is given, then there exists dðaÞ > 0 such that if, x!c, , 0 < jx  cj < dðaÞ and x 2 A, then f ðxÞ > a. But since f ðxÞ  gðxÞ for all x 2 A; x 6¼ c,, it follows that if 0 < jx  cj < dðaÞ and x 2 A, then gðxÞ > a. Therefore lim g ¼ 1., x!c, , The proof of (b) is similar., , Q.E.D., , The function gðxÞ ¼ 1=x considered in Example 4.3.6(b) suggests that it might be, useful to consider one-sided infinite limits. We will define only right-hand infinite, limits., 4.3.8 Definition Let A
R and let f : A ! R. If c 2 R is a cluster point of the set, A \ ðc; 1Þ ¼ fx 2 A : x > cg, then we say that f tends to 1 [respectively, 1] as, x ! cþ, and we write, h, i, lim f ¼ 1 respectively; lim f ¼ 1 ;, x!cþ, , x!cþ, , if for every a 2 R there is d ¼ dðaÞ > 0 such that for all x 2 A with 0 < x  c < d, then, f ðxÞ > a [respectively, f ðxÞ < a]., 4.3.9 Examples (a) Let gðxÞ :¼ 1=x for x 6¼ 0. We have noted in Example 4.3.6(b) that, lim g does not exist. However, it is an easy exercise to show that, , x!0, , lim ð1=xÞ ¼ 1 and, , x!0þ, , lim ð1=xÞ ¼ 1:, , x!0, , (b) It was seen in Example 4.3.4(b) that the function gðxÞ :¼ e1=x for x 6¼ 0 is not, bounded on any interval ð0; dÞ; d > 0. Hence the right-hand limit of e1=x as x ! 0þ does, not exist in the sense of Definition 4.3.1(i). However, since, 1=x < e1=x, , for, , x > 0;, , it is readily seen that lim e1=x ¼ 1 in the sense of Definition 4.3.8., x!0þ, , &, , Limits at Infinity, It is also desirable to define the notion of the limit of a function as x ! 1. The definition as, x ! 1 is similar.

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C04, , 12/09/2010, , 14:49:16, , Page 121, , 4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT, , 121, , 4.3.10 Definition Let A
R and let f : A ! R. Suppose that ða; 1Þ
A for some, a 2 R. We say that L 2 R is a limit of f as x ! 1, and write, lim f ¼ L or, , x!1, , lim f ðxÞ ¼ L ;, , x!1, , if given any e > 0 there exists K ¼ KðeÞ > a such that for any x > K, then j f ðxÞ  Lj < e., The reader should note the close resemblance between 4.3.10 and the definition of a, limit of a sequence., We leave it to the reader to show that the limits of f as x ! 1 are unique whenever, they exist. We also have sequential criteria for these limits; we shall only state the criterion, as x ! 1. This uses the notion of the limit of a properly divergent sequence (see, Definition 3.6.1)., 4.3.11 Theorem Let A
R, let f : A ! R, and suppose that ða; 1Þ
A for some, a 2 R. Then the following statements are equivalent:, (i) L ¼ lim f ., x!1, (ii) For every sequence (xn) in A \ ða; 1Þ such that limðxn Þ ¼ 1, the sequence ð f ðxn ÞÞ, converges to L., We leave it to the reader to prove this theorem and to formulate and prove the, companion result concerning the limit as x ! 1., 4.3.12 Examples (a) Let gðxÞ :¼ 1=x for x 6¼ 0., It is an elementary exercise to show that lim ð1=xÞ ¼ 0 ¼ lim ð1=xÞ. (See Figure, x!1, x!1, 4.3.4.), (b) Let f ðxÞ :¼ 1=x2 for x 6¼ 0., The reader may show that lim ð1=x2 Þ ¼ 0 ¼ lim ð1=x2 Þ. (See Figure 4.3.3.) One, x!1, , x!1, , way to do this is to show that if x  1 then 0  1=x2  1=x. In view of part (a), this implies, that lim ð1=x2 Þ ¼ 0., &, x!1, , Just as it is convenient to be able to say that f ðxÞ ! 1 as x ! c for c 2 R, it is, convenient to have the corresponding notion as x ! 1. We will treat the case where, x ! 1., 4.3.13 Definition Let A
R and let f : A ! R. Suppose that ða; 1Þ
A for some, a 2 A. We say that f tends to 1 [respectively, 1] as x ! 1, and write, h, i, lim f ¼ 1 respectively; lim f ¼ 1, x!1, , x!1, , if given any a 2 R there exists K ¼ KðaÞ > a such that for any x > K, then f ðxÞ > a, [respectively, f ðxÞ < a]., As before there is a sequential criterion for this limit., 4.3.14 Theorem Let A 2 R, let f : A ! R, and suppose that ða; 1Þ
A for some, a 2 R. Then the following statements are equivalent:

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C04, , 12/09/2010, , 14:49:16, , 122, (i), , Page 122, , CHAPTER 4 LIMITS, , lim f ¼ 1 ½respectively; lim f ¼ 1., , x!1, , x!1, , (ii) For every sequence (xn) in ða; 1Þsuch that limðxn Þ ¼ 1, then limð f ðxn ÞÞ ¼ 1, ½respectively; limð f ðxn ÞÞ ¼ 1., The next result is an analogue of Theorem 3.6.5., 4.3.15 Theorem Let A
R, let f ; g : A ! R, and suppose that ða; 1Þ
A for some, a 2 R. Suppose further that gðxÞ > 0 for all x > a and that for some L 2 R; L 6¼ 0, we have, lim, , f ðxÞ, , x!1 gðxÞ, , ¼ L:, , (i), , If L > 0, then lim f ¼ 1 if and only if lim g ¼ 1., , (ii), , If L < 0, then lim f ¼ 1 if and only if lim g ¼ 1., , x!1, , x!1, , x!1, , Proof., , (i), , x!1, , Since L > 0, the hypothesis implies that there exists a1 > a such that, , 1, f ðxÞ 3, < L, for, x > a1 :, 0< L, 2, gðxÞ 2,  ,  , Therefore we have 12 L gðxÞ < f ðxÞ < 32 L gðxÞ for all x > a1, from which the conclusion, follows readily., The proof of (ii) is similar., Q.E.D., We leave it to the reader to formulate the analogous result as x ! 1., 4.3.16 Examples (a) lim xn ¼ 1 for n 2 N., x!1, , Let gðxÞ :¼ xn for x 2 ð0; 1Þ. Given a 2 R, let K :¼ supf1; ag. Then for all x > K,, we have gðxÞ ¼ xn  x > a. Since a 2 R is arbitrary, it follows that lim g ¼ 1., x!1, , lim xn ¼ 1 for n 2 N, n even, and lim xn ¼ 1 for n 2 N, n odd., x!1, x!1, We will treat the case n odd, say n ¼ 2k þ 1 with k ¼ 0, 1, . . . . Given a 2 R, let, K :¼ inffa; 1g. For any x < K, then since ðx2 Þk  1, we have xn ¼ ðx2 Þk x  x < a., Since a 2 R is arbitrary, it follows that lim xn ¼ 1., , (b), , x!1, , (c) Let p : R ! R be the polynomial function, pðxÞ :¼ an xn þ an1 xn1 þ    þ a1 x þ a0 :, Then lim p ¼ 1 if an > 0, and lim p ¼ 1 if an < 0., x!1, , x!1, , Indeed, let gðxÞ :¼ xn and apply Theorem 4.3.15. Since, , , pðxÞ, 1, 1, ¼ an þ an1, þ    þ a1 n1, gðxÞ, x, x, , , þ a0, , 1, ;, xn, , it follows that lim ðpðxÞ=gðxÞÞ ¼ an . Since lim g ¼ 1, the assertion follows from, x!1, x!1, Theorem 4.3.15., (d) Let p be the polynomial function in part (c). Then lim p ¼ 1 [respectively, 1] if, x!1, n is even [respectively, odd] and an > 0., &, We leave the details to the reader.

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C05, , 12/08/2010, , 14:19:37, , Page 124, , CHAPTER 5, , CONTINUOUS FUNCTIONS, , We now begin the study of the most important class of functions that arises in real analysis:, the class of continuous functions. The term ‘‘continuous’’ has been used since the time of, Newton to refer to the motion of bodies or to describe an unbroken curve, but it was not made, precise until the nineteenth century. Work of Bernhard Bolzano in 1817 and Augustin-Louis, Cauchy in 1821 identified continuity as a very significant property of functions and proposed, definitions, but since the concept is tied to that of limit, it was the careful work of Karl, Weierstrass in the 1870s that brought proper understanding to the idea of continuity., We will first define the notions of continuity at a point and continuity on a set, and then, show that various combinations of continuous functions give rise to continuous functions., Then in Section 5.3 we establish the fundamental properties that make continuous functions, so important. For instance, we will prove that a continuous function on a closed bounded, interval must attain a maximum and a minimum value. We also prove that a continuous, function must take on every value intermediate to any two values it attains. These properties, and others are not possessed by general functions, as various examples illustrate, and thus, they distinguish continuous functions as a very special class of functions., In Section 5.4 we introduce the very important notion of uniform continuity. The, distinction between continuity and uniform continuity is somewhat subtle and was not fully, appreciated until the work of Weierstrass and the mathematicians of his era, but it proved to, be very significant in applications. We present one application to the idea of approximating, continuous functions by more elementary functions (such as polynomials)., , Karl Weierstrass, Karl Weierstrass (¼Weierstrab) (1815–1897) was born in Westphalia,, Germany. His father, a customs officer in a salt works, insisted that he study, law and public finance at the University of Bonn, but he had more interest in, drinking and fencing, and left Bonn without receiving a diploma. He then, enrolled in the Academy of M€unster where he studied mathematics with, Christoph Gudermann. From 1841 to 1854 he taught at various gymnasia in, Prussia. Despite the fact that he had no contact with the mathematical world, during this time, he worked hard on mathematical research and was able to, publish a few papers, one of which attracted considerable attention. Indeed,, the University of K€onigsberg gave him an honorary doctoral degree for this work in 1855. The next, year, he secured positions at the Industrial Institute of Berlin and the University of Berlin. He, remained at Berlin until his death., A methodical and painstaking scholar, Weierstrass distrusted intuition and worked to put, everything on a firm and logical foundation. He did fundamental work on the foundations of, arithmetic and analysis, on complex analysis, the calculus of variations, and algebraic geometry., Due to his meticulous preparation, he was an extremely popular lecturer; it was not unusual for, him to speak about advanced mathematical topics to audiences of more than 250. Among his, auditors are counted Georg Cantor, Sonya Kovalevsky, G€, osta Mittag-Leffler, Max Planck, Otto, H€older, David Hilbert, and Oskar Bolza (who had many American doctoral students). Through his, writings and his lectures, Weierstrass had a profound influence on contemporary mathematics., , 124

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C05, , 12/08/2010, , 14:19:38, , Page 125, , 5.1 CONTINUOUS FUNCTIONS, , 125, , The notion of a ‘‘gauge’’ is introduced in Section 5.5 and is used to provide an alternative, method of proving the fundamental properties of continuous functions. The main significance of this concept, however, is in the area of integration theory where gauges are essential, in defining the generalized Riemann integral. This will be discussed in Chapter 10., Monotone functions are an important class of functions with strong continuity, properties and they are discussed in Section 5.6., , Section 5.1 Continuous Functions, In this section, which is very similar to Section 4.1, we will define what it means to say that, a function is continuous at a point, or on a set. This notion of continuity is one of the central, concepts of mathematical analysis, and it will be used in almost all of the following, material in this book. Consequently, it is essential that the reader master it., 5.1.1 Definition Let A  R, let f : A ! R, and let c 2 A. We say that f is continuous at, c if, given any number e > 0, there exists d > 0 such that if x is any point of A satisfying, jx  cj < d, then j f ðxÞ  f ðcÞj < e., If f fails to be continuous at c, then we say that f is discontinuous at c., As with the definition of limit, the definition of continuity at a point can be formulated, very nicely in terms of neighborhoods. This is done in the next result. We leave the, verification as an important exercise for the reader. See Figure 5.1.1., , Figure 5.1.1, , Given V e ð f ðcÞÞ, a neighborhood V d ðcÞ is to be determined, , 5.1.2 Theorem A function f : A ! R is continuous at a point c 2 A if and only if given, any e-neighborhood V e ð f ðcÞÞ of f (c) there exists a d-neighborhood V d ðcÞ of c such that if, x is any point of A \ V d ðcÞ, then f (x) belongs to V e ð f ðcÞÞ, that is,, f ðA \ V d ðcÞÞ  V e ð f ðcÞÞ:, Remarks (1) If c 2 A is a cluster point of A, then a comparison of Definitions 4.1.4, and 5.1.1 show that f is continuous at c if and only if, ð1Þ, , f ðcÞ ¼ lim f ðxÞ:, x!c

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C05, , 12/08/2010, , 14:19:39, , 126, , Page 126, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Thus, if c is a cluster point of A, then three conditions must hold for f to be continuous at c:, (i) f must be defined at c (so that f (c) makes sense),, (ii) the limit of f at c must exist in R (so that lim f ðxÞ makes sense), and, x!c, (iii) these two values must be equal., (2) If c 2 A is not a cluster point of A, then there exists a neighborhood V d ðcÞ of c such, that A \ V d ðcÞ ¼ fcg. Thus we conclude that a function f is automatically continuous at a, point c 2 A that is not a cluster point of A. Such points are often called ‘‘isolated points’’ of, A. They are of little practical interest to us, since they have no relation to a limiting process., Since continuity is automatic for such points, we generally test for continuity only at cluster, points. Thus we regard condition (1) as being characteristic for continuity at c., A slight modification of the proof of Theorem 4.1.8 for limits yields the following, sequential version of continuity at a point., 5.1.3 Sequential Criterion for Continuity A function f : A ! R is continuous at the, point c 2 A if and only if for every sequence (xn) in A that converges to c, the sequence, ð f ðxn ÞÞ converges to f (c)., The following Discontinuity Criterion is a consequence of the last theorem. It should, be compared with the Divergence Criterion 4.1.9(a) with L ¼ f (c). Its proof should be, written out in detail by the reader., 5.1.4 Discontinuity Criterion Let A  R, let f : A ! R, and let c 2 A. Then f is, discontinuous at c if and only if there exists a sequence (xn) in A such that (xn) converges, to c, but the sequence ð f ðxn ÞÞ does not converge to f (c)., So far we have discussed continuity at a point. To talk about the continuity of a, function on a set, we will simply require that the function be continuous at each point of the, set. We state this formally in the next definition., 5.1.5 Definition Let A  R and let f : A ! R. If B is a subset of A, we say that f is, continuous on the set B if f is continuous at every point of B., 5.1.6 Examples (a) The constant function f ðxÞ :¼ b is continuous on R., It was seen in Example 4.1.7(a) that if c 2 R, then lim f ðxÞ ¼ b. Since f ðcÞ ¼ b, we, x!c, have lim f ðxÞ ¼ f ðcÞ, and thus f is continuous at every point c 2 R. Therefore f is, x!c, , continuous on R., (b) gðxÞ :¼ x is continuous on R., It was seen in Example 4.1.7(b) that if c 2 R, then we have lim g ¼ c. Since gðcÞ ¼ c,, x!c, then g is continuous at every point c 2 R. Thus g is continuous on R., (c) hðxÞ :¼ x2 is continuous on R., It was seen in Example 4.1.7(c) that if c 2 R, then we have lim h ¼ c2 . Since, x!c, hðcÞ ¼ c2 , then h is continuous at every point c 2 R. Thus h is continuous on R., (d) wðxÞ :¼ 1=x is continuous on A :¼ fx 2 R : x > 0g., It was seen in Example 4.1.7(d) that if c 2 A, then we have lim w ¼ 1=c. Since, x!c, wðcÞ ¼ 1=c, this shows that w is continuous at every point c 2 A. Thus w is continuous on A.

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C05, , 12/08/2010, , 14:19:39, , Page 127, , 5.1 CONTINUOUS FUNCTIONS, , 127, , (e) wðxÞ :¼ 1=x is not continuous at x ¼ 0., Indeed, if wðxÞ ¼ 1=x for x > 0, then w is not defined for x ¼ 0, so it cannot be, continuous there. Alternatively, it was seen in Example 4.1.10(a) that lim w does not exist, x!0, in R, so w cannot be continuous at x ¼ 0., (f) The signum function sgn is not continuous at 0., The signum function was defined in Example 4.1.10(b), where it was also shown that, lim sgnðxÞ does not exist in R. Therefore sgn is not continuous at x ¼ 0 (even though sgn 0, , x!0, , is defined). It is an exercise to show that sgn is continuous at every point c 6¼ 0., , Note In the next two examples, we introduce functions that played a significant role in, the development of real analysis. Discontinuities are emphasized and it is not possible to, graph either of them satisfactorily. The intuitive idea of drawing a curve in the plane to, represent a function simply does not apply, and plotting a handful of points gives only a hint, of their character. In the nineteenth century, these functions clearly demonstrated the need, for a precise and rigorous treatment of the basic concepts of analysis. They will reappear in, later sections., (g) Let A :¼ R and let f be Dirichlet’s ‘‘discontinuous function’’ defined by,
, f ðxÞ :¼, , 1 if, 0 if, , x is rational;, x is irrational:, , We claim that f is not continuous at any point of R. (This function was introduced in 1829, by P. G. L. Dirichlet.), Indeed, if c is a rational number, let (xn) be a sequence of irrational numbers that, converges to c. (Corollary 2.4.9 to the Density Theorem 2.4.8 assures us that such a, sequence does exist.) Since f ðxn Þ ¼ 0 for all n 2 N, we have limð f ðxn ÞÞ ¼ 0, while, f ðcÞ ¼ 1. Therefore f is not continuous at the rational number c., On the other hand, if b is an irrational number, let ( yn) be a sequence of rational, numbers that converge to b. (The Density Theorem 2.4.8 assures us that such a sequence, does exist.) Since f ðyn Þ ¼ 1 for all n 2 N, we have limð f ð yn ÞÞ ¼ 1, while f ðbÞ ¼ 0., Therefore f is not continuous at the irrational number b., Since every real number is either rational or irrational, we deduce that f is not, continuous at any point in R., , Figure 5.1.2, , Thomae’s function

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C05, , 12/08/2010, , 14:19:40, , 128, , Page 128, , CHAPTER 5 CONTINUOUS FUNCTIONS, , (h) Let A :¼ fx 2 R : x > 0g. For any irrational number x > 0 we define hðxÞ :¼ 0., For a rational number in A of the form m=n, with natural numbers m, n having no, common factors except 1, we define hðm=nÞ :¼ 1=n. (We also define hð0Þ :¼ 1.), We claim that h is continuous at every irrational number in A, and is discontinuous at, every rational number in A. (This function was introduced in 1875 by K. J. Thomae.), Indeed, if a > 0 is rational, let (xn) be a sequence of irrational numbers in, A that converges to a. Then limðhðxn ÞÞ ¼ 0, while h(a) > 0. Hence h is discontinuous, at a., On the other hand, if b is an irrational number and e > 0, then (by the Archimedean, Property) there is a natural number n0 such that 1=n0 < e. There are only a finite number, of rationals with denominator less than n0 in the interval ðb  1; b þ 1Þ. (Why?) Hence, d > 0 can be chosen so small that the neighborhood ðb  d; b þ dÞ contains no rational, numbers with denominator less than n0. It then follows that for jx  bj < d; x 2 A,, we have jhðxÞ  hðbÞj ¼ jhðxÞj  1=n0 < e. Thus h is continuous at the irrational, number b., Consequently, we deduce that Thomae’s function h is continuous precisely at the, &, irrational points in A. (See Figure 5.1.2.), 5.1.7 Remarks (a) Sometimes a function f : A ! R is not continuous at a point c, because it is not defined at this point. However, if the function f has a limit L at the point c, and if we define F on A [ fcg ! R by,
, FðxÞ :¼, , L, f ðxÞ, , for, for, , x ¼ c;, x 2 A;, , then F is continuous at c. To see this, one needs to check that lim F ¼ L, but this follows, x!c, (why?), since lim f ¼ L., x!c, (b) If a function g : A ! R does not have a limit at c, then there is no way that we can, obtain a function G : A [ fcg ! R that is continuous at c by defining,
, GðxÞ :¼, , C, gðxÞ, , for, for, , x ¼ c;, x 2 A:, , To see this, observe that if lim G exists and equals C, then lim g must also exist and, x!c, x!c, equal C., 5.1.8 Examples (a) The function gðxÞ :¼ sinð1=xÞ for x 6¼ 0 (see Figure 4.1.3) does, not have a limit at x ¼ 0 (see Example 4.1.10(c)). Thus there is no value that we can assign, at x ¼ 0 to obtain a continuous extension of g at x ¼ 0., (b) Let f ðxÞ :¼ x sin ð1=xÞ for x 6¼ 0. (See Figure 5.1.3.) It was seen in Example 4.2.8(f), that lim ðx sinð1=xÞÞ ¼ 0. Therefore it follows from Remark 5.1.7(a) that if we define, x!0, , F : R ! R by,
, FðxÞ :¼, then F is continuous at x ¼ 0., , 0, for, x sinð1=xÞ for, , x ¼ 0;, x 6¼ 0;, &

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C05, , 12/08/2010, , 14:19:41, , Page 129, , 5.1 CONTINUOUS FUNCTIONS, , Figure 5.1.3, , 129, , Graph of f ðxÞ ¼ x sinð1=xÞ ðx 6¼ 0Þ, , Exercises for Section 5.1, 1. Prove the Sequential Criterion 5.1.3., 2. Establish the Discontinuity Criterion 5.1.4., 3. Let a < b < c. Suppose that f is continuous on [a, b], that g is continuous on [b, c], and that, f ðbÞ ¼ gðbÞ. Define h on [a, c] by hðxÞ :¼ f ðxÞ for x 2 ½a; b and hðxÞ :¼ gðxÞ for x 2 ½b; c., Prove that h is continuous on [a, c]., 4. If x 2 R, we define vxb to be the greatest integer n 2 Z such that n  x. (Thus, for example,, v8:3b ¼ 8; vpb ¼ 3; v  pb ¼ 4.) The function x 7! vxb is called the greatest integer function., Determine the points of continuity of the following functions:, (a) f ðxÞ :¼ vxb;, (b) gðxÞ :¼ x vxb;, (c) hðxÞ :¼ vsin xb;, (d) kðxÞ :¼ v1=xb ðx 6¼ 0Þ:, 5. Let f be defined for all x 2 R; x 6¼ 2, by f ðxÞ ¼ ðx2 þ x  6Þ=ðx  2Þ. Can f be defined at, x ¼ 2 in such a way that f is continuous at this point?, 6. Let A  R and let f : A ! R be continuous at a point c 2 A. Show that for any e > 0, there exists, a neighborhood V d ðcÞ of c such that if x; y 2 A \ V d ðcÞ, then j f ðxÞ  f ðyÞj < e., 7. Let f : R ! R be continuous at c and let f ðcÞ > 0. Show that there exists a neighborhood V d ðcÞ, of c such that if x 2 V d ðcÞ, then f ðxÞ > 0., 8. Let f : R ! R be continuous on R and let S :¼ fx 2 R : f ðxÞ ¼ 0g be the ‘‘zero set’’ of f. If, ðxn Þ is in S and x ¼ limðxn Þ, show that x 2 S., 9. Let A  B  R, let f : B ! R and let g be the restriction of f to A (that is, gðxÞ ¼ f ðxÞ for, x 2 A)., (a) If f is continuous at c 2 A, show that g is continuous at c., (b) Show by example that if g is continuous at c, it need not follow that f is continuous, at c., 10. Show that the absolute value function f ðxÞ :¼ jxj is continuous at every point c 2 R., 11. Let K > 0 and let f : R ! R satisfy the condition j f ðxÞ  f ðyÞj  Kjx  yj for all x; y 2 R., Show that f is continuous at every point c 2 R., 12. Suppose that f : R ! R is continuous on R and that f ðrÞ ¼ 0 for every rational number r. Prove, that f ðxÞ ¼ 0 for all x 2 R., 13. Define g : R ! R by gðxÞ :¼ 2x for x rational, and gðxÞ :¼ x þ 3 for x irrational. Find all, points at which g is continuous.

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C05, , 12/08/2010, , 14:19:42, , 130, , Page 130, , CHAPTER 5 CONTINUOUS FUNCTIONS, , 14. Let A :¼ ð0; 1Þ and let k : A ! R be defined as follows. For x 2 A; x irrational, we define, kðxÞ ¼ 0; for x 2 A rational and of the form x ¼ m=n with natural numbers m, n having no, common factors except 1, we define kðxÞ :¼ n. Prove that k is unbounded on every open interval, in A. Conclude that k is not continuous at any point of A. (See Example 5.1.6(h).), 15. Let f : ð0; 1Þ ! R be bounded but such that lim f does not exist. Show that there are two, x!0, sequences ðxn Þ and ðyn Þ in (0, 1) with limðxn Þ ¼ 0 ¼ limðyn Þ, but such that ð f ðxn ÞÞ and ð f ðyn ÞÞ, exist but are not equal., , Section 5.2 Combinations of Continuous Functions, Let A  R and let f and g be functions that are defined on A to R and let b 2 R. In Definition, 4.2.3 we defined the sum, difference, product, and multiple functions denoted by, f þ g; f  g; f g; bf . In addition, if h : A ! R is such that hðxÞ 6¼ 0 for all x 2 A, then, we defined the quotient function denoted by f =h., The next result is similar to Theorem 4.2.4, from which it follows., 5.2.1 Theorem Let A  R, let f and g be functions on A to R, and let b 2 R. Suppose, that c 2 A and that f and g are continuous at c., (a) Then f þ g; f  g; f g, and bf are continuous at c., (b) If h : A ! R is continuous at c 2 A and if hðxÞ 6¼ 0 for all x 2 A, then the quotient, f=h is continuous at c., Proof. If c 2 A is not a cluster point of A, then the conclusion is automatic. Hence we, assume that c is a cluster point of A., (a) Since f and g are continuous at c, then, f ðcÞ ¼ lim f and gðcÞ ¼ lim g:, x!c, , x!c, , Hence it follows from Theorem 4.2.4(a) that, ð f þ gÞðcÞ ¼ f ðcÞ þ gðcÞ ¼ lim ð f þ gÞ:, x!c, , Therefore f þ g is continuous at c. The remaining assertions in part (a) are proved in a, similar fashion., (b) Since c 2 A, then hðcÞ 6¼ 0. But since hðcÞ ¼ lim h, it follows from Theorem 4.2.4(b), x!c, that,  , lim f, f, f ðcÞ x!c, f, ðcÞ ¼, ¼, ¼ lim, :, h, hðcÞ lim h x!c h, x!c, , Therefore f=h is continuous at c., , Q.E.D., , The next result is an immediate consequence of Theorem 5.2.1, applied to every point, of A. However, since it is an extremely important result, we shall state it formally., 5.2.2 Theorem Let A  R, let f and g be continuous on A to R, and let b 2 R., (a) The functions f þ g; f  g; f g, and bf are continuous on A., (b) If h : A ! R is continuous on A and hðxÞ 6¼ 0 for x 2 A, then the quotient f=h is, continuous on A.

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C05, , 12/08/2010, , 14:19:43, , Page 131, , 5.2 COMBINATIONS OF CONTINUOUS FUNCTIONS, , 131, , Remark To define quotients, it is sometimes more convenient to proceed as follows. If, w : A ! R, let A1 :¼ fx 2 A : wðxÞ 6¼ 0g. We can define the quotient f =w on the set A1 by, ð1Þ, ,  , f, f ðxÞ, ðxÞ :¼, w, w ðxÞ, , for, , x 2 A1 :, , If w is continuous at a point c 2 A1 , it is clear that the restriction w1 of w to A1 is also, continuous at c. Therefore it follows from Theorem 5.2.1(b) applied to w1 that f =w1, is continuous at c 2 A. Since ð f =wÞðxÞ ¼ ð f =w1 ÞðxÞ for x 2 A1 it follows that f =w is, continuous at c 2 A1 . Similarly, if f and w are continuous on A, then the function f =w,, defined on A1 by (1), is continuous on A1., 5.2.3 Examples (a) Polynomial functions., If p is a polynomial function, so that pðxÞ ¼ an xn þ an1 xn1 þ    þ a1 x þ a0 for all, x 2 R, then it follows from Example 4.2.5(f) that pðcÞ ¼ lim p for any c 2 R. Thus a, x!c, polynomial function is continuous on R., (b) Rational functions., If p and q are polynomial functions on R, then there are at most a finite number, a1 ; . . . ; am of real roots of q. If x 2, = fa1 ; . . . ; am g then qðxÞ 6¼ 0 so that we can define the, rational function r by, rðxÞ :¼, , pðxÞ, qðxÞ, , for, , x2, = fa1 ; . . . ; am g :, , It was seen in Example 4.2.5(g) that if qðcÞ 6¼ 0, then, rðcÞ ¼, , pðcÞ, pðxÞ, ¼ lim, ¼ lim rðxÞ:, qðcÞ x!c qðxÞ x!c, , In other words, r is continuous at c. Since c is any real number that is not a root of q, we, infer that a rational function is continuous at every real number for which it is defined., (c) We shall show that the sine function sin is continuous on R., To do so we make use of the following properties of the sine and cosine functions., (See Section 8.4.) For all x, y, z 2 R we have:, jsin zj  jzj; jcos zj  1;, ,  , , sin x  sin y ¼ 2 sin 12ðx  yÞ cos 12ðx þ yÞ :, Hence if c 2 R, then we have, jsin x  sin cj  2  12 jx  cj  1 ¼ jx  cj:, Therefore sin is continuous at c. Since c 2 R is arbitrary, it follows that sin is continuous, on R., (d) The cosine function is continuous on R., We make use of the following properties of the sine and cosine functions. For all, x; y; z 2 R we have:, jsin zj  jzj; jsin zj  1;, ,  , , cos x  cos y ¼ 2 sin 12ðx þ yÞ sin 12ðx  yÞ :, Hence if c 2 R, then we have, jcos x  cos cj  2  1  12jc  xj ¼ jx  cj:

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C05, , 12/08/2010, , 14:19:43, , 132, , Page 132, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Therefore cos is continuous at c. Since c 2 R is arbitrary, it follows that cos is continuous, on R. (Alternatively, we could use the relation cos x ¼ sinðx þ p=2Þ.), (e) The functions tan, cot, sec, csc are continuous where they are defined., For example, the cotangent function is defined by, cos x, cot x :¼, sin x, provided sin x 6¼ 0 (that is, provided x 6¼ np; n 2 Z). Since sin and cos are continuous on, R, it follows (see the Remark before Example 5.2.3) that the function cot is continuous, &, on its domain. The other trigonometric functions are treated similarly., 5.2.4 Theorem Let A  R, let f : A ! R, and let j f j be defined by j f jðxÞ :¼ j f ðxÞj for, x 2 A., (a) If f is continuous at a point c 2 A, then j f j is continuous at c., (b) If f is continuous on A, then j f j is continuous on A., Proof. This is an immediate consequence of Exercise 4.2.14., 5.2.5 Theorem Let Ap, A !ffi R, and let f ðxÞ  0 for all x 2 A. We let, ffiffiffi R, let fp: ffiffiffiffiffiffiffiffi, defined for x 2 A by ð f ÞðxÞ :¼ f ðxÞ., pffiffiffi, (a) If f is continuous at a point c 2 A, then f is continuous at c., pffiffiffi, (b) If f is continuous on A, then f is continuous on A., Proof. This is an immediate consequence of Exercise 4.2.15., , Q.E.D., , pffiffiffi, f be, , Q.E.D., , Composition of Continuous Functions, We now show that if the function f : A ! R is continuous at a point c and if g : B ! R is, continuous at b ¼ f (c), then the composition g  f is continuous at c. In order to assure that, g  f is defined on all of A, we also need to assume that f ðAÞ  B., 5.2.6 Theorem Let A, B  R and let f : A ! R and g : B ! R be functions such that, f ðAÞ  B. If f is continuous at a point c 2 A and g is continuous at b ¼ f ðcÞ 2 B, then the, composition g  f : A ! R is continuous at c., Proof. Let W be an e-neighborhood of g(b). Since g is continuous at b, there is a, d-neighborhood V of b ¼ f ðcÞ such that if y 2 B \ V then gðyÞ 2 W. Since f is continuous, at c, there is a g-neighborhood U of c such that if x 2 A \ U, then f ðxÞ 2 V. (See, Figure 5.2.1.) Since f ðAÞ  B, it follows that if x 2 A \ U, then f ðxÞ 2 B \ V so that, g  f ðxÞ ¼ gð f ðxÞÞ 2 W. But since W is an arbitrary e-neighborhood of g(b), this implies, that g  f is continuous at c., Q.E.D., 5.2.7 Theorem Let A; B  R, let f : A ! R be continuous on A, and let g : B ! R be, continuous on B. If f ðAÞ  B, then the composite function g  f : A ! R is continuous on A., Proof. The theorem follows immediately from the preceding result, if f and g are, continuous at every point of A and B, respectively., Q.E.D.

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C05, , 12/08/2010, , 14:19:44, , Page 133, , 5.2 COMBINATIONS OF CONTINUOUS FUNCTIONS, , Figure 5.2.1, , 133, , The composition of f and g, , Theorems 5.2.6 and 5.2.7 are very useful in establishing that certain functions are, continuous. They can be used in many situations where it would be difficult to apply the, definition of continuity directly., 5.2.8 Examples (a) Let g1 ðxÞ :¼ jxj for x 2 R. It follows from the Triangle Inequality, that, jg1 ðxÞ  g1 ðcÞj  jx  cj, for all x; c 2 R. Hence g1 is continuous at c 2 R. If f : A ! R is any function that is, continuous on A, then Theorem 5.2.7 implies that g1  f ¼ j f j is continuous on A. This, gives another proof of Theorem 5.2.4., pffiffiffi, (b) Let g2 ðxÞ :¼ x for x  0. It follows from Theorems 3.2.10 and 5.1.3 that g2 is, continuous at any number c  0. If f : A ! R is continuous, pffiffiffi on A and if f ðxÞ  0 for all, x 2 A, then it follows from Theorem 5.2.7 that g2  f ¼ f is continuous on A. This gives, another proof of Theorem 5.2.5., (c) Let g3 ðxÞ :¼ sin x for x 2 R. We have seen in Example 5.2.3(c) that g3 is continuous, on R. If f : A ! R is continuous on A, then it follows from Theorem 5.2.7 that g3  f is, continuous on A., In particular, if f ðxÞ :¼ 1=x for x 6¼ 0, then the function gðxÞ :¼ sinð1=xÞ is continuous at every point c 6¼ 0. [We have seen, in Example 5.1.8(a), that g cannot be defined at, &, 0 in order to become continuous at that point.], , Exercises for Section 5.2, 1. Determine the points of continuity of the following functions and state which theorems are used, in each case., pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, x2 þ 2x þ 1, pffiffiffiffi, ðx 2 R Þ;, (b) gðxÞ :¼ x þ x ðx  0Þ;, (a) f ðxÞ :¼, x2 þ 1, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, pffiffiffiffiffiffiffiffiffiffiffiffiffi, (d) kðxÞ :¼ cos 1 þ x2 ðx 2 R Þ:, (c) hðxÞ :¼ 1 þ j sin xj ðx 6¼ 0Þ;, x, 2. Show that if f : A ! R is continuous on A  R and if n 2 N, then the function f n defined by, f n ðxÞ ¼ ð f ðxÞÞn , for x 2 A, is continuous on A., 3. Give an example of functions f and g that are both discontinuous at a point c in R such that (a) the, sum f þ g is continuous at c, (b) the product fg is continuous at c.

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C05, , 12/08/2010, , 14:19:45, , 134, , Page 134, , CHAPTER 5 CONTINUOUS FUNCTIONS, , 4. Let x 7! vxb denote the greatest integer function (see Exercise 5.1.4). Determine the points of, continuity of the function f ðxÞ :¼ x  vxb; x 2 R., 5. Let g be defined on R by gð1Þ :¼ 0, and gðxÞ :¼ 2 if x 6¼ 1, and let f ðxÞ :¼ x þ 1 for all x 2 R., Show that lim g  f 6¼ ðg  f Þð0Þ. Why doesn’t this contradict Theorem 5.2.6?, x!0, , 6. Let f, g be defined on R and let c 2 R. Suppose that lim f ¼ b and that g is continuous at b., x!c, Show that lim g  f ¼ gðbÞ. (Compare this result with Theorem 5.2.7 and the preceding, x!c, , exercise.), 7. Give an example of a function f : ½0; 1 ! R that is discontinuous at every point of [0, 1] but, such that j f j is continuous on [0, 1]., 8. Let f, g be continuous from R to R, and suppose that f ðrÞ ¼ gðrÞ for all rational numbers r. Is it, true that f ðxÞ ¼ gðxÞ for all x 2 R?, 9. Let h : R ! R be continuous on R satisfying hðm=2n Þ ¼ 0 for all m 2 Z; n 2 N. Show that, hðxÞ ¼ 0 for all x 2 R., 10. Let f : R ! R be continuous on R, and let P :¼ fx 2 R : f ðxÞ > 0g. If c 2 P, show that there, exists a neighborhood V d ðcÞ  P., 11. If f and g are continuous on R, let S :¼ fx 2 R : f ðxÞ  gðxÞg. If ðsn Þ  S and limðsn Þ ¼ s,, show that s 2 S., 12. A function f : R ! R is said to be additive if f ðx þ yÞ ¼ f ðxÞ þ f ðyÞ for all x, y in R. Prove, that if f is continuous at some point x0, then it is continuous at every point of R. (See, Exercise 4.2.12.), 13. Suppose that f is a continuous additive function on R. If c :¼ f ð1Þ, show that we have f ðxÞ ¼ cx, for all x 2 R. [Hint: First show that if r is a rational number, then f ðrÞ ¼ cr.], 14. Let g : R ! R satisfy the relation gðx þ yÞ ¼ gðxÞ gðyÞ for all x, y in R. Show that if g is, continuous at x ¼ 0, then g is continuous at every point of R. Also if we have gðaÞ ¼ 0 for some, a 2 R, then gðxÞ ¼ 0 for all x 2 R., 15. Let f ; g : R ! R be continuous at a point c, and let hðxÞ :¼ supf f ðxÞ; gðxÞg for x 2 R. Show, that hðxÞ ¼ 12 ð f ðxÞ þ gðxÞÞ þ 12 j f ðxÞ  gðxÞj for all x 2 R. Use this to show that h is, continuous at c., , Section 5.3 Continuous Functions on Intervals, Functions that are continuous on intervals have a number of very important properties that, are not possessed by general continuous functions. In this section, we will establish some, deep results that are of considerable importance and that will be applied later. Alternative, proofs of these results will be given in Section 5.5., 5.3.1 Definition A function f : A ! R is said to be bounded on A if there exists a, constant M > 0 such that j f ðxÞj  M for all x 2 A., In other words, a function is bounded on a set if its range is a bounded set in R. To say, that a function is not bounded on a given set is to say that no particular number can serve, as a bound for its range. In exact language, a function f is not bounded on the set A if given, any M > 0, there exists a point xM 2 A such that j f ðxM Þj > M. We often say that f is, unbounded on A in this case., For example, the function f defined on the interval A :¼ ð0; 1Þ by f ðxÞ :¼ 1=x is not, bounded on A because for any M > 0 we can take the point xM :¼ 1=ðM þ 1Þ in A to get, f ðxM Þ ¼ 1=xM ¼ M þ 1 > M. This example shows that continuous functions need not be

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C05, , 12/08/2010, , 14:19:45, , Page 135, , 5.3 CONTINUOUS FUNCTIONS ON INTERVALS, , 135, , bounded. In the next theorem, however, we show that continuous functions on a certain, type of interval are necessarily bounded., 5.3.2 Boundedness Theoremy Let I :¼ ½a; b be a closed bounded interval and let, f : I ! R be continuous on I. Then f is bounded on I., Proof. Suppose that f is not bounded on I. Then, for any n 2 N there is a number xn 2 I, such that j f ðxn Þj > n. Since I is bounded, the sequence X :¼ ðxn Þ is bounded. Therefore,, the Bolzano-Weierstrass Theorem 3.4.8 implies that there is a subsequence X 0 ¼ ðxnr Þ of X, that converges to a number x. Since I is closed and the elements of X 0 belong to I, it follows, from Theorem 3.2.6 that x 2 I. Then f is continuous at x, so that ð f ðxnr ÞÞ converges to f ðxÞ., We then conclude from Theorem 3.2.2 that the convergent sequence ðf ðxnr ÞÞ must be, bounded. But this is a contradiction since, j f ðxnr Þj > nr  r for, , r 2 N:, , Therefore the supposition that the continuous function f is not bounded on the closed, bounded interval I leads to a contradiction., Q.E.D., To show that each hypothesis of the Boundedness Theorem is needed, we can, construct examples that show the conclusion fails if any one of the hypotheses is relaxed., (i) The interval must be bounded. The function f ðxÞ :¼ x for x in the unbounded,, closed interval A :¼ ½0; 1Þ is continuous but not bounded on A., (ii) The interval must be closed. The function gðxÞ :¼ 1=x for x in the half-open, interval B :¼ ð0; 1 is continuous but not bounded on B., (iii) The function must be continuous. The function h defined on the closed interval, C :¼ ½0; 1 by hðxÞ :¼ 1=x for x 2 ð0; 1 and hð0Þ :¼ 1 is discontinuous and unbounded, on C., The Maximum-Minimum Theorem, 5.3.3 Definition Let A  R and let f : A ! R. We say that f has an absolute maximum, on A if there is a point x 2 A such that, f ðx Þ  f ðxÞ for all x 2 A:, We say that f has an absolute minimum on A if there is a point x 2 A such that, f ðx Þ  f ðxÞ, , for all, , x 2 A:, , We say that x is an absolute maximum point for f on A, and that x is an absolute, minimum point for f on A, if they exist., We note that a continuous function on a set A does not necessarily have an absolute, maximum or an absolute minimum on the set. For example, f ðxÞ :¼ 1=x has neither an, absolute maximum nor an absolute minimum on the set A :¼ ð0; 1Þ. (See Figure 5.3.1.), There can be no absolute maximum for f on A since f is not bounded above on A, and there, is no point at which f attains the value 0 ¼ inff f ðxÞ : x 2 Ag. The same function has, y, , This theorem, as well as 5.3.4, is true for an arbitrary closed bounded set. For these developments, see Sections, 11.2 and 11.3.

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C05, , 12/08/2010, , 14:19:45, , 136, , Page 136, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Figure 5.3.1 The function, f ðxÞ ¼ 1=x ðx > 0Þ, , Figure 5.3.2, gðxÞ ¼ x2, , The function, ðjxj  1Þ, , neither an absolute maximum nor an absolute minimum when it is restricted to the set, (0, 1), while it has both an absolute maximum and an absolute minimum when it is, restricted to the set [1, 2]. In addition, f ðxÞ ¼ 1=x has an absolute maximum but no, absolute minimum when restricted to the set [1, 1), but no absolute maximum and no, absolute minimum when restricted to the set (1, 1)., It is readily seen that if a function has an absolute maximum point, then this point is, not necessarily uniquely determined. For example, the function gðxÞ :¼ x2 defined for, x 2 A :¼ ½1; þ1 has the two points x ¼ 1 giving the absolute maximum on A, and the, single point x ¼ 0 yielding its absolute minimum on A. (See Figure 5.3.2.) To pick an, extreme example, the constant function hðxÞ :¼ 1 for x 2 R is such that every point of R is, both an absolute maximum and an absolute minimum point for h., 5.3.4 Maximum-Minimum Theorem Let I :¼ ½a; b be a closed bounded interval and, let f : I ! R be continuous on I. Then f has an absolute maximum and an absolute, minimum on I., Proof. Consider the nonempty set f ðIÞ :¼ ff ðxÞ : x 2 Ig of values of f on I. In Theorem 5.3.2, it was established that f (I ) is a bounded subset of R. Let s :¼ sup f ðIÞ and s :¼ inf f ðI Þ., We claim that there exist points x and x in I such that s ¼ f ðx Þ and s ¼ f ðx Þ. We will, establish the existence of the point x , leaving the proof of the existence of x to the reader., Since s ¼ sup f ðIÞ, if n 2 N, then the number s  1=n is not an upper bound of the, set f (I). Consequently there exists a number xn 2 I such that, ð1Þ, , 1, s  < f ðxn Þ  s, n, , for all, , n 2 N:, , Since I is bounded, the sequence X :¼ ðxn Þ is bounded. Therefore, by the BolzanoWeierstrass Theorem 3.4.8, there is a subsequence X 0 ¼ ðxnr Þ of X that converges to some, number x . Since the elements of X0 belong to I ¼ ½a; b, it follows from Theorem 3.2.6 that, x 2 I. Therefore f is continuous at x so that limð f ðxnr ÞÞ ¼ f ðx Þ. Since it follows from, (1) that, s , , 1, < f ðxnr Þ  s, nr, , for all, , r 2 N;, , we conclude from the Squeeze Theorem 3.2.7 that limð f ðxnr ÞÞ ¼ s . Therefore we have, f ðx Þ ¼ limð f ðxnr ÞÞ ¼ s ¼ sup f ðIÞ:, We conclude that x is an absolute maximum point of f on I., , Q.E.D.

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C05, , 12/08/2010, , 14:19:48, , Page 137, , 5.3 CONTINUOUS FUNCTIONS ON INTERVALS, , 137, , The next result is the theoretical basis for locating roots of a continuous function by, means of sign changes of the function. The proof also provides an algorithm, known as the, Bisection Method, for the calculation of roots to a specified degree of accuracy and can be, readily programmed for a computer. It is a standard tool for finding solutions of equations, of the form f (x) ¼ 0, where f is a continuous function. An alternative proof of the theorem, is indicated in Exercise 5.3.11., 5.3.5 Location of Roots Theorem Let I ¼ ½a; b and let f : I ! R be continuous on I. If, f ðaÞ < 0 < f ðbÞ, or if f ðaÞ > 0 > f ðbÞ, then there exists a number c 2 ða; bÞ such that, f ðcÞ ¼ 0., Proof. We assume that f ðaÞ < 0 < f ðbÞ. We will generate a sequence of intervals by, successive bisections. Let I 1 :¼ ½a1 ; b1 , where a1 :¼ a; b1 :¼ b, and let p1 be the midpoint, p1 :¼ 12 ða1 þ b1 Þ. If f ðp1 Þ ¼ 0, we take c :¼ p1 and we are done. If f ðp1 Þ 6¼ 0, then either, f ðp1 Þ > 0 or f ðp1 Þ < 0. If f ðp1 Þ > 0, then we set a2 :¼ a1 ; b2 :¼ p1 , while if f ðp1 Þ < 0,, then we set a2 :¼ p1 ; b2 :¼ b1 . In either case, we let I 2 :¼ ½a2 ; b2 ; then we have I 2 I 1 and, f ða2 Þ < 0; f ðb2 Þ > 0., We continue the bisection process. Suppose that the intervals I 1 ; I 2 ; . . . ; I k have, been obtained by successive bisection in the same manner. Then we have f ðak Þ < 0 and, f ðbk Þ > 0, and we set pk :¼ 12 ðak þ bk Þ. If f ðpk Þ ¼ 0, we take c :¼ pk and we are done., If f ðpk Þ > 0, we set akþ1 :¼ ak ; bkþ1 :¼ pk , while if f ðpk Þ < 0, we set akþ1 :¼, pk ; bkþ1 :¼ bk . In either case, we let I kþ1 :¼ ½akþ1 ; bkþ1 ; then I kþ1 I k and, f ðakþ1 Þ < 0; f ðbkþ1 Þ > 0., If the process terminates by locating a point pn such that f ðpn Þ ¼ 0, then we are done., If the process does not terminate, then we obtain a nested sequence of closed bounded, intervals I n :¼ ½an ; bn  such that for every n 2 N we have, f ðan Þ < 0, , and f ðbn Þ > 0:, , Furthermore, since the intervals are obtained by repeated bisection, the length of In is, equal to bn  an ¼ ðb  aÞ=2n1 . It follows from the Nested Intervals Property 2.5.2, that there exists a point c that belongs to In for all n 2 N. Since an  c  bn for all, n 2 N and limðbn  an Þ ¼ 0, it follows that limðan Þ ¼ c ¼ limðbn Þ. Since f is continuous, at c, we have, lim ð f ðan ÞÞ ¼ f ðcÞ ¼ lim ð f ðbn ÞÞ:, The fact that f ðan Þ < 0 for all n 2 N implies that f ðcÞ ¼ lim ð f ðan ÞÞ  0. Also, the fact, that f ðbn Þ > 0 for all n 2 N implies that f ðcÞ ¼ lim ð f ðbn ÞÞ  0. Thus, we conclude that, f ðcÞ ¼ 0. Consequently, c is a root of f., Q.E.D., The following example illustrates how the Bisection Method for finding roots is, applied in a systematic fashion., 5.3.6 Example The equation f ðxÞ ¼ xex  2 ¼ 0 has a root c in the interval [0, 1],, because f is continuous on this interval and f ð0Þ ¼ 2 < 0 and f ð1Þ ¼ e  2 > 0. Using a, calculator we construct the following table, where the sign of f ( pn) determines the interval, at the next step. The far right column is an upper bound on the error when pn is used to, approximate the root c, because we have, jpn  cj  12 ðbn  an Þ ¼ 1=2n :

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C05, , 12/08/2010, , 14:19:49, , 138, , Page 138, , CHAPTER 5 CONTINUOUS FUNCTIONS, , We will find an approximation pn with error less than 102., n, 1, 2, 3, 4, 5, 6, 7, , an, 0, .5, .75, .75, .8125, .84375, .84375, , bn, 1, 1, 1, .875, .875, .875, .859375, , pn, .5, .75, .875, .8125, .84375, .859375, .8515625, , f ( pn), 1.176, .412, þ.099, .169, .0382, þ.0296, —, , 1, 2 ðbn, ,  an Þ, .5, .25, .125, .0625, .03125, .015625, .0078125, , We have stopped at n ¼ 7, obtaining c p7 ¼ :8515625 with error less than .0078125., This is the first step in which the error is less than 102. The decimal place values of, p7 past the second place cannot be taken seriously, but we can conclude that, &, :843 < c < :860., Bolzano’s Theorem, The next result is a generalization of the Location of Roots Theorem. It assures us that a, continuous function on an interval takes on (at least once) any number that lies between, two of its values., 5.3.7 Bolzano’s Intermediate Value Theorem Let I be an interval and let f : I ! R be, continuous on I. If a; b 2 I and if k 2 R satisfies f ðaÞ < k < f ðbÞ, then there exists a point, c 2 I between a and b such that f ðcÞ ¼ k., Proof. Suppose that a < b and let gðxÞ :¼ f ðxÞ  k; then gðaÞ < 0 < gðbÞ. By the Location, of Roots Theorem 5.3.5 there exists a point c with a < c < b such that 0 ¼ gðcÞ ¼ f ðcÞ  k., Therefore f ðcÞ ¼ k., If b < a, let hðxÞ :¼ k  f ðxÞ so that hðbÞ < 0 < hðaÞ. Therefore there exists a point c, with b < c < a such that 0 ¼ hðcÞ ¼ k  f ðcÞ, whence f ðcÞ ¼ k., Q.E.D., 5.3.8 Corollary Let I ¼ ½a; b be a closed, bounded interval and let f : I ! R be, continuous on I. If k 2 R is any number satisfying, inf f ðIÞ  k  sup f ðIÞ;, then there exists a number c 2 I such that f ðcÞ ¼ k., Proof. It follows from the Maximum-Minimum Theorem 5.3.4 that there are points c, and c in I such that, inf f ðIÞ ¼ f ðc Þ  k  f ðc Þ ¼ sup f ðIÞ:, The conclusion now follows from Bolzano’s Theorem 5.3.7., , Q.E.D., , The next theorem summarizes the main results of this section. It states that the image, of a closed bounded interval under a continuous function is also a closed bounded interval., The endpoints of the image interval are the absolute minimum and absolute maximum, values of the function, and the statement that all values between the absolute minimum, and the absolute maximum values belong to the image is a way of describing Bolzano’s, Intermediate Value Theorem.

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C05, , 12/08/2010, , 14:19:50, , Page 139, , 5.3 CONTINUOUS FUNCTIONS ON INTERVALS, , 139, , 5.3.9 Theorem Let I be a closed bounded interval and let f : I ! R be continuous on I., Then the set f ðIÞ :¼ f f ðxÞ : x 2 Ig is a closed bounded interval., Proof. If we let m :¼ inf f ðIÞ and M :¼ sup f ðIÞ, then we know from the MaximumMinimum Theorem 5.3.4 that m and M belong to f (I). Moreover, we have f ðIÞ  ½m; M. If, k is any element of [m, M], then it follows from the preceding corollary that there exists a, point c 2 I such that k ¼ f ðcÞ. Hence, k 2 f ðIÞ and we conclude that ½m; M  f ðIÞ., Therefore, f (I) is the interval [m, M]., Q.E.D., Warning If I :¼ ½a; b is an interval and f : I ! R is continuous on I, we have proved, that f (I) is the interval [m, M]. We have not proved (and it is not always true) that f (I) is the, &, interval ½ f ðaÞ; f ðbÞ. (See Figure 5.3.3.), , Figure 5.3.3, , f ðIÞ ¼ ½m; M, , The preceding theorem is a ‘‘preservation’’ theorem in the sense that it states that the, continuous image of a closed bounded interval is a set of the same type. The next theorem, extends this result to general intervals. However, it should be noted that although the, continuous image of an interval is shown to be an interval, it is not true that the image, interval necessarily has the same form as the domain interval. For example, the continuous, image of an open interval need not be an open interval, and the continuous image of an, unbounded closed interval need not be a closed interval. Indeed, if f ðxÞ :¼ 1=ðx2 þ 1Þ for, x 2 R, then f is continuous on R [see Example 5.2.3(b)]. It is easy to see that if, I 1 :¼ ð1; 1Þ, then f ðI 1 Þ ¼ 12 ; 1 , which is not an open interval. Also, if I 2 :¼ ½0; 1Þ,, then f ðI 2 Þ ¼ ð0; 1, which is not a closed interval. (See Figure 5.3.4.), , Figure 5.3.4, , Graph of f ðxÞ ¼ 1=ðx2 þ 1Þ ðx 2 RÞ

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C05, , 12/08/2010, , 14:19:53, , 140, , Page 140, , CHAPTER 5 CONTINUOUS FUNCTIONS, , To prove the Preservation of Intervals Theorem 5.3.10, we will use Theorem 2.5.1, characterizing intervals., 5.3.10 Preservation of Intervals Theorem Let I be an interval and let f : I ! R be, continuous on I. Then the set f (I) is an interval., Proof. Let a; b 2 f ðIÞ with a < b; then there exist points a; b 2 I such that a ¼ f ðaÞ, and b ¼ f ðbÞ. Further, it follows from Bolzano’s Intermediate Value Theorem 5.3.7, that if k 2 ða; bÞ then there exists a number c 2 I with k ¼ f ðcÞ 2 f ðIÞ. Therefore, ½a; b  f ðIÞ, showing that f (I ) possesses property (1) of Theorem 2.5.1. Therefore, f (I ) is an interval., Q.E.D., Exercises for Section 5.3, 1. Let I :¼ ½a; b and let f : I ! R be a continuous function such that f ðxÞ > 0 for each x in I., Prove that there exists a number a > 0 such that f ðxÞ  a for all x 2 I., 2. Let I :¼ ½a; b and let f : I ! R and g : I ! R be continuous functions on I. Show that the set, E :¼ fx 2 I : f ðxÞ ¼ gðxÞg has the property that if ðxn Þ  E and xn ! x0 , then x0 2 E., 3. Let I :¼ ½a; b and let f : I ! R be a continuous function on I such that for each x in I there, exists y in I such that j f ðyÞj  12 j f ðxÞj. Prove there exists a point c in I such that f ðcÞ ¼ 0., 4. Show that every polynomial of odd degree with real coefficients has at least one real root., 5. Show that the polynomial pðxÞ :¼ x4 þ 7x3  9 has at least two real roots. Use a calculator to, locate these roots to within two decimal places., 6. Let f be continuous on the interval [0, 1] to R and such that f ð0Þ ¼ f ð1Þ. Prove,  that there exists, a point c in [0, 12] such that f ðcÞ ¼ f c þ 12 . [Hint: Consider gðxÞ ¼ f ðxÞ  f x þ 12 .] Conclude, that there are, at any time, antipodal points on the earth’s equator that have the same, temperature., 7. Show that the equation x ¼ cos x has a solution in the interval ½0; p=2. Use the Bisection, Method and a calculator to find an approximate solution of this equation, with error less than, 103., pffiffiffi, 8. Show that the function f ðxÞ :¼ 2 ln x þ x  2 has root in the interval [1, 2], Use the Bisection, Method and a calculator to find the root with error less than 102., 9. (a) The function f ðxÞ :¼ ðx  1Þðx  2Þðx  3Þðx  4Þðx  5Þ has five roots in the interval, [0, 7]. If the Bisection Method is applied on this interval, which of the roots is located?, (b) Same question for gðxÞ :¼ ðx  2Þðx  3Þðx  4Þðx  5Þðx  6Þ on the interval [0, 7]., 10. If the Bisection Method is used on an interval of length 1 to find pn with error j pn  cj < 105,, determine the least value of n that will assure this accuracy., 11. Let I :¼ ½a; b, let f : I ! R be continuous on I, and assume that f ðaÞ < 0; f ðbÞ > 0. Let, W :¼ fx 2 I : f ðxÞ < 0g, and let w :¼ sup W. Prove that f ðwÞ ¼ 0. (This provides an alternative proof of Theorem 5.3.5.), 12. Let I :¼ ½0; p=2 and let f : I ! R be defined by f ðxÞ :¼ supfx2 ; cos xg for x 2 I. Show there, exists an absolute minimum point x0 2 I for f on I. Show that x0 is a solution to the equation, cos x ¼ x2 ., 13. Suppose that f : R ! R is continuous on R and that lim f ¼ 0 and lim f ¼ 0. Prove that f is, x!1, x!1, bounded on R and attains either a maximum or minimum on R. Give an example to show that, both a maximum and a minimum need not be attained., 14. Let f : R ! R be continuous on R and let b 2 R. Show that if x0 2 R is such that f ðx0 Þ < b,, then there exists a d-neighborhood U of x0 such that f ðxÞ < b for all x 2 U.

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C05, , 12/08/2010, , 14:19:54, , Page 141, , 5.4 UNIFORM CONTINUITY, , 141, , 15. Examine which open [respectively, closed] intervals are mapped by f ðxÞ :¼ x2 for x 2 R onto, open [respectively, closed] intervals., 16. Examine the mapping of open [respectively, closed] intervals under the functions gðxÞ :¼, 1=ðx2 þ 1Þ and hðxÞ :¼ x3 for x 2 R., 17. If f : ½0; 1 ! R is continuous and has only rational [respectively, irrational] values, must f be, constant? Prove your assertion., 18. Let I :¼ ½a; b and let f : I ! R be a (not necessarily continuous) function with the property that, for every x 2 I, the function f is bounded on a neighborhood V dx ðxÞ of x (in the sense of, Definition 4.2.1). Prove that f is bounded on I., 19. Let J :¼ ða; bÞ and let g : J ! R be a continuous function with the property that for every x 2 J,, the function g is bounded on a neighborhood V dx ðxÞ of x. Show by example that g is not, necessarily bounded on J., , Section 5.4 Uniform Continuity, Let A  R and let f : A ! R. Definition 5.1.1 states that the following statements are, equivalent:, (i) f is continuous at every point u 2 A;, (ii) given e > 0 and u 2 A, there is a dðe; uÞ > 0 such that for all x such that x 2 A, and jx  uj < dðe; uÞ, then j f ðxÞ  f ðuÞj < e., The point we wish to emphasize here is that d depends, in general, on both e > 0 and, u 2 A. The fact that d depends on u is a reflection of the fact that the function f may change, its values rapidly near certain points and slowly near other points. [For example, consider, f ðxÞ :¼ sinð1=xÞ for x > 0; see Figure 4.1.3.], Now it often happens that the function f is such that the number d can be chosen to be, independent of the point u 2 A and to depend only on e. For example, if f ðxÞ :¼ 2x for all, x 2 R, then, j f ðxÞ  f ðuÞj ¼ 2jx  uj;, and so we can choose dðe; uÞ :¼ e=2 for all e > 0 and all u 2 R. (Why?), On the other hand if gðxÞ :¼ 1=x for x 2 A :¼ fx 2 R : x > 0g, then, ux, :, ð1Þ, gðxÞ  gðuÞ ¼, ux, If u 2 A is given and if we take, ð2Þ, , dðe; uÞ :¼ inf, , 1, 1 2, 2 u; 2 u e, , ;, , then if jx  uj < dðe; uÞ, we have jx  uj < 12 u so that 12 u < x < 32 u, whence it follows that, 1=x < 2=u. Thus, if jx  uj < 12 u, the equality (1) yields the inequality, ð3Þ, , jgðxÞ  gðuÞj  ð2=u2 Þjx  uj:, , Consequently, if jx  uj < dðe; uÞ, then (2) and (3) imply that, , jgðxÞ  gðuÞj < ð2=u2 Þ 12 u2 e ¼ e:, We have seen that the selection of dðe; uÞ by the formula (2) ‘‘works’’ in the sense that it, enables us to give a value of d that will ensure that jgðxÞ  gðuÞj < e when jx  uj < d and

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C05, , 12/08/2010, , 14:19:55, , 142, , Page 142, , CHAPTER 5 CONTINUOUS FUNCTIONS, , x; u 2 A. We note that the value of dðe; uÞ given in (2) certainly depends on the point u 2 A., If we wish to consider all u 2 A, formula (2) does not lead to one value dðeÞ > 0 that will, ‘‘work’’ simultaneously for all u > 0, since inffdðe; uÞ : u > 0g ¼ 0., In fact, there is no way of choosing one value of d that will ‘‘work’’ for all u > 0 for the, function gðxÞ ¼ 1=x. The situation isexhibited graphically in Figures 5.4.1 and,  5.4.2, where, for a given e-neighborhood V e 12 about 12 ¼ f ð2Þ and V e ð2Þ about 2 ¼ f 12 , the, corresponding maximum values of d are seen to be considerably different. As u tends to 0,, the permissible values of d tend to 0., , Figure 5.4.1, , gðxÞ ¼ 1=x ðx > 0Þ, , Figure 5.4.2, , gðxÞ ¼ 1=x ðx > 0Þ, , 5.4.1 Definition Let A  R and let f : A ! R. We say that f is uniformly continuous, on A if for each e > 0 there is a dðeÞ > 0 such that if x; u 2 A are any numbers satisfying, jx  uj < dðeÞ, then j f ðxÞ  f ðuÞj < e., It is clear that if f is uniformly continuous on A, then it is continuous at every point of A., In general, however, the converse does not hold, as is shown by the function gðxÞ ¼ 1=x on, the set A :¼ fx 2 R : x > 0g., It is useful to formulate a condition equivalent to saying that f is not uniformly, continuous on A. We give such criteria in the next result, leaving the proof to the reader as, an exercise., 5.4.2 Nonuniform Continuity Criteria Let A  R and let f : A ! R. Then the following statements are equivalent:, (i), (ii), (iii), , f is not uniformly continuous on A., There exists an e0 > 0 such that for every d > 0 there are points xd ; ud in A such that, jxd  ud j < d and j f ðxd Þ  f ðud Þj  e0 ., There exists an e0 > 0 and two sequences (xn) and (un) in A such that limðxn , un Þ ¼ 0 and j f ðxn Þ  f ðun Þj  e0 for all n 2 N., , We can apply this result to show that gðxÞ :¼ 1=x is not uniformly continuous, on A :¼ fx 2 R : x > 0g. For, if xn :¼ 1=n and un :¼ 1=ðn þ 1Þ, then we have, limðxn  un Þ ¼ 0, but jgðxn Þ  gðun Þj ¼ 1 for all n 2 N., We now present an important result that assures that a continuous function on a closed, bounded interval I is uniformly continuous on I. Other proofs of this theorem are given in, Sections 5.5 and 11.3.

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C05, , 12/08/2010, , 14:19:57, , Page 143, , 5.4 UNIFORM CONTINUITY, , 143, , 5.4.3 Uniform Continuity Theorem Let I be a closed bounded interval and let, f : I ! R be continuous on I. Then f is uniformly continuous on I., Proof. If f is not uniformly continuous on I then, by the preceding result, there exists, e0 > 0 and two sequences (xn) and (un) in I such that jxn  un j < 1=n and j f ðxn Þ  f ðun Þj , e0 for all n 2 N. Since I is bounded, the sequence (xn) is bounded; by the BolzanoWeierstrass Theorem 3.4.8 there is a subsequence ðxnk Þ of (xn) that converges to an element, z. Since I is closed, the limit z belongs to I, by Theorem 3.2.6. It is clear that the, corresponding subsequence ðunk Þ also converges to z, since, junk  zj  junk  xnk j þ jxnk  zj:, Now if f is continuous at the point z, then both of the sequences ð f ðxnk ÞÞ and ð f ðunk ÞÞ, must converge to f ðzÞ. But this is not possible since, j f ðxn Þ  f ðun Þj  e0, for all n 2 N. Thus the hypothesis that f is not uniformly continuous on the closed bounded, interval I implies that f is not continuous at some point z 2 I. Consequently, if f is, continuous at every point of I, then f is uniformly continuous on I., Q.E.D., Lipschitz Functions, If a uniformly continuous function is given on a set that is not a closed bounded interval,, then it is sometimes difficult to establish its uniform continuity. However, there is a, condition that frequently occurs that is sufficient to guarantee uniform continuity. It is, named after Rudolf Lipschitz (1832–1903) who was a student of Dirichlet and who worked, extensively in differential equations and Riemannian geometry., 5.4.4 Definition Let A  R and let f : A ! R. If there exists a constant K > 0 such that, ð4Þ, , j f ðxÞ  f ðuÞj  Kjx  uj, , for all x; u 2 A, then f is said to be a Lipschitz function (or to satisfy a Lipschitz, condition) on A., The condition (4) that a function f : I ! R on an interval I is a Lipschitz function can, be interpreted geometrically as follows. If we write the condition as, f ðxÞ  f ðuÞ,  K;, xu, , x; u 2 I; x 6¼ u ;, , then the quantity inside the absolute values is the slope of a line segment joining the points, ðx; f ðxÞÞ and ðu; f ðuÞÞ. Thus a function f satisfies a Lipschitz condition if and only if the, slopes of all line segments joining two points on the graph of y ¼ f ðxÞ over I are bounded, by some number K., 5.4.5 Theorem If f : A ! R is a Lipschitz function, then f is uniformly continuous on A., Proof. If condition (4) is satisfied, then given e > 0, we can take d :¼ e=K. If x; u 2 A, satisfy jx  uj < d, then, e, j f ðxÞ  f ðuÞj < K  ¼ e:, K, Therefore f is uniformly continuous on A., Q.E.D.

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C05, , 12/08/2010, , 14:19:58, , 144, , Page 144, , CHAPTER 5 CONTINUOUS FUNCTIONS, , 5.4.6 Examples (a) If f ðxÞ :¼ x2 on A :¼ ½0; b, where b > 0, then, j f ðxÞ  f ðuÞj ¼ jx þ ujjx  uj  2bjx  uj, for all x, u in [0, b]. Thus f satisfies (4) with K :¼ 2b on A, and therefore f is uniformly, continuous on A. Of course, since f is continuous and A is a closed bounded interval, this, can also be deduced from the Uniform Continuity Theorem. (Note that f does not satisfy a, Lipschitz condition on the interval ½0; 1Þ.), (b) Not every uniformly, continuous function is a Lipschitz function., pffiffiffi, Let gðxÞ :¼ x for x in the closed bounded interval I :¼ ½0; 2. Since g is continuous, on I, it follows from the Uniform Continuity Theorem 5.4.3 that g is uniformly continuous, on I. However, there is no number K > 0 such that jgðxÞj  Kjxj for all x 2 I. (Why not?), Therefore, g is not a Lipschitz function on I., (c) The Uniform Continuity Theorem and Theorem 5.4.5 can sometimes be combined to, establish the uniform continuity, pffiffiffi of a function on a set., We consider gðxÞ :¼ x on the set A :¼ ½0; 1Þ. The uniform continuity of g on the, interval I :¼ ½0; 2 follows from the Uniform Continuity Theorem as noted in (b). If, J :¼ ½1; 1Þ, then if both x, u are in J, we have, jgðxÞ  gðuÞj ¼, , pffiffiffi pffiffiffi, jx  uj, x  u ¼ pffiffiffi pffiffiffi  12 jx  uj:, xþ u, , Thus g is a Lipschitz function on J with constant K ¼ 12, and hence by Theorem 5.4.5, g is, uniformly continuous on [1, 1). Since A ¼ I [ J, it follows [by taking dðeÞ :¼, inf f1; dI ðeÞ; dJ ðeÞg] that g is uniformly continuous on A. We leave the details to the, &, reader., , The Continuous Extension Theorem, We have seen examples of functions that are continuous but not uniformly continuous on, open intervals; for example, the function f ðxÞ ¼ 1=x on the interval (0, 1). On the other, hand, by the Uniform Continuity Theorem, a function that is continuous on a closed, bounded interval is always uniformly continuous. So the question arises: Under what, conditions is a function uniformly continuous on a bounded open interval? The answer, reveals the strength of uniform continuity, for it will be shown that a function on (a, b) is, uniformly continuous if and only if it can be defined at the endpoints to produce a function, that is continuous on the closed interval. We first establish a result that is of interest in itself., 5.4.7 Theorem If f : A ! R is uniformly continuous on a subset A of R and if ðxn Þ is a, Cauchy sequence in A, then ð f ðxn ÞÞ is a Cauchy sequence in R., Proof. Let (xn) be a Cauchy sequence in A, and let e > 0 be given. First choose d > 0 such, that if x, u in A satisfy jx  uj < d, then j f ðxÞ  f ðuÞj < e. Since (xn) is a Cauchy, sequence, there exists HðdÞ such that jxn  xm j < d for all n; m > HðdÞ. By the choice of d,, this implies that for n; m > HðdÞ, we have j f ðxn Þ  f ðxm Þj < e. Therefore the sequence, ð f ðxn ÞÞ is a Cauchy sequence., Q.E.D., The preceding result gives us an alternative way of seeing that f ðxÞ :¼ 1=x is not, uniformly continuous on (0, 1). We note that the sequence given by xn :¼ 1=n in (0, 1) is a, Cauchy sequence, but the image sequence, where f ðxn Þ ¼ n, is not a Cauchy sequence.

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C05, , 12/08/2010, , 14:19:59, , Page 145, , 5.4 UNIFORM CONTINUITY, , 145, , 5.4.8 Continuous Extension Theorem A function f is uniformly continuous on the, interval (a, b) if and only if it can be defined at the endpoints a and b such that the extended function is continuous on [a, b]., Proof. ð(Þ This direction is trivial., ð)Þ Suppose f is uniformly continuous on (a, b). We shall show how to extend f to a;, the argument for b is similar. This is done by showing that x!c, lim f ðxÞ ¼ L exists, and this is, accomplished by using the sequential criterion for limits. If (xn) is a sequence in (a, b) with, limðxn Þ ¼ a, then it is a Cauchy sequence, and by the preceding theorem, the sequence, ð f ðxn ÞÞ is also a Cauchy sequence, and so is convergent by Theorem 3.5.5. Thus the limit, limð f ðxn ÞÞ ¼ L exists. If (un) is any other sequence in (a, b) that converges to a, then, limðun  xn Þ ¼ a  a ¼ 0, so by the uniform continuity of f we have, limð f ðun ÞÞ ¼ limð f ðun Þ  f ðxn ÞÞ þ limð f ðxn ÞÞ, ¼ 0 þ L ¼ L:, Since we get the same value L for every sequence converging to a, we infer from the, sequential criterion for limits that f has limit L at a. If we define f ðaÞ :¼ L, then f is, continuous at a. The same argument applies to b, so we conclude that f has a continuous, extension to the interval [a, b]., Q.E.D., Since the limit of f ðxÞ :¼ sinð1=xÞ at 0 does not exist, we infer from the Continuous, Extension Theorem that the function is not uniformly continuous on ð0; b for any b > 0., On the other hand, since lim x sinð1=xÞ ¼ 0 exists, the function gðxÞ :¼ x sinð1=xÞ is, x!0, , uniformly continuous on ð0; b for all b > 0., Approximationy, In many applications it is important to be able to approximate continuous functions by, functions of an elementary nature. Although there are a variety of definitions that can be used, to make the word ‘‘approximate’’ more precise, one of the most natural (as well as one of the, most important) is to require that, at every point of the given domain, the approximating, function shall not differ from the given function by more than the preassigned error., 5.4.9 Definition A function s : ½a; b ! R is called a step function if [a, b] is the union, of a finite number of nonoverlapping intervals I 1 ; I 2 ; . . . ; I n such that s is constant on each, interval, that is, sðxÞ ¼ ck for all x 2 I k ; k ¼ 1; 2; . . . ; n., Thus a step function has only a finite number of distinct values., For example, the function s : ½2; 4 ! R defined by, 8, 0; 2  x < 1;, >, >, >, >, 1; 1  x  0;, >, >, < 1, 0 < x < 12 ;, 2;, sðxÞ :¼, 1, 3;, >, >, 2  x < 1;, >, >, 2;, 1,  x  3;, >, >, :, 2;, 3 < x  4;, is a step function. (See Figure 5.4.3.), , y, , The rest of this section can be omitted on a first reading of this chapter.

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C05, , 12/08/2010, , 14:20:0, , 146, , Page 146, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Figure 5.4.3, , Graph of y ¼ s(x), , We will now show that a continuous function on a closed bounded interval I can be, approximated arbitrarily closely by step functions., 5.4.10 Theorem Let I be a closed bounded interval and let f : I ! R be continuous on I., If e > 0, then there exists a step function se : I ! R such that j f ðxÞ  se ðxÞj < e for all, x 2 I., Proof. Since (by the Uniform Continuity Theorem 5.4.3) the function f is uniformly, continuous, it follows that given e > 0 there is a number dðeÞ > 0 such that if x; y 2 I and, jx  yj < dðeÞ, then j f ðxÞ  f ð yÞj < e. Let I :¼ ½a; b and let m 2 N be sufficiently large so, that h :¼ ðb  aÞ=m < dðeÞ. We now divide I ¼ [a, b] into m disjoint intervals of length h;, namely, I 1 :¼ ½a; a þ h, and I k :¼ ða þ ðk  1Þh; a þ khÞ for k ¼ 2, . . . , m. Since the, length of each subinterval Ik is h < dðeÞ, the difference between any two values of f in Ik is, less than e. We now define, ð5Þ, , se ðxÞ :¼ f ða þ khÞ for, , x 2 Ik;, , k ¼ 1; . . . ; m;, , so that se is constant on each interval Ik. (In fact the value of se on Ik is the value of f at the, right endpoint of Ik. See Figure 5.4.4.) Consequently if x 2 I k , then, j f ðxÞ  se ðxÞj ¼ j f ðxÞ  f ða þ khÞj < e:, Therefore we have j f ðxÞ  se ðxÞj < e for all x 2 I., , Figure 5.4.4, , Approximation by step functions, , Q.E.D.

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C05, , 12/08/2010, , 14:20:2, , Page 147, , 5.4 UNIFORM CONTINUITY, , 147, , Note that the proof of the preceding theorem establishes somewhat more than was, announced in the statement of the theorem. In fact, we have proved the following, more, precise, assertion., 5.4.11 Corollary Let I :¼ ½a; b be a closed bounded interval and let f : I ! R be, continuous on I. If e > 0, there exists a natural number m such that if we divide I into m, disjoint intervals Ik having length h :¼ ðb  aÞ=m, then the step function se defined in, equation (5) satisfies j f ðxÞ  se ðxÞj < e for all x 2 I., Step functions are extremely elementary in character, but they are not continuous, (except in trivial cases). Since it is often desirable to approximate continuous functions by, elementary continuous functions, we now shall show that we can approximate continuous, functions by continuous piecewise linear functions., 5.4.12 Definition Let I :¼ ½a; b be an interval. Then a function g : I ! R is said to be, piecewise linear on I if I is the union of a finite number of disjoint intervals I 1 ; . . . ; I m ,, such that the restriction of g to each interval Ik is a linear function., Remark It is evident that in order for a piecewise linear function g to be continuous on, I, the line segments that form the graph of g must meet at the endpoints of adjacent, subintervals I k ; I kþ1 ðk ¼ 1; . . . ; m  1Þ., 5.4.13 Theorem Let I be a closed bounded interval and let f : I ! R be continuous on I., If e > 0, then there exists a continuous piecewise linear function ge : I ! R such that, j f ðxÞ  ge ðxÞj < e for all x 2 I., Proof. Since f is uniformly continuous on I :¼ ½a; b, there is a number dðeÞ > 0 such that, if x, y 2 I and jx  yj < dðeÞ, then j f ðxÞ  f ðyÞj < e. Let m 2 N be sufficiently large so, that h :¼ ðb  aÞ=m < dðeÞ. Divide I ¼ ½a; b into m disjoint intervals of length h; namely,, let I 1 ¼ ½a; a þ h, and let I k ¼ ða þ ðk  1Þh; a þ kh for k ¼ 2; . . . ; m. On each interval, Ik we define ge to be the linear function joining the points, ða þ ðk  1Þh; f ða þ ðk  1ÞhÞÞ and, , ða þ kh; f ða þ khÞÞ :, , Then ge is a continuous piecewise linear function on I. Since, for x 2 I k the value f(x) is, within e of f ða þ ðk  1ÞhÞ and f ða þ khÞ, it is an exercise to show that j f ðxÞ  ge ðxÞj < e, for all x 2 I k ; therefore this inequality holds for all x 2 I. (See Figure 5.4.5.), Q.E.D., , Figure 5.4.5, , Approximation by piecewise linear function

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C05, , 12/08/2010, , 14:20:3, , 148, , Page 148, , CHAPTER 5 CONTINUOUS FUNCTIONS, , We shall close this section by stating the important theorem of Weierstrass concerning the approximation of continuous functions by polynomial functions. As would be, expected, in order to obtain an approximation within an arbitrarily preassigned e > 0,, we must be prepared to use polynomials of arbitrarily high degree., 5.4.14 Weierstrass Approximation Theorem Let I ¼ ½a; b and let f : I ! R be a, continuous function. If e > 0 is given, then there exists a polynomial function pe such that, j f ðxÞ  pe ðxÞj < e for all x 2 I., There are a number of proofs of this result. Unfortunately, all of them are rather, intricate, or employ results that are not yet at our disposal. (A proof can be found in Bartle,, ERA, pp. 169–172, which is listed in the References.), , Exercises for Section 5.4, 1. Show that the function f ðxÞ :¼ 1=x is uniformly continuous on the set A :¼ ½a; 1Þ, where a is a, positive constant., 2. Show that the function f ðxÞ :¼ 1=x2 is uniformly continuous on A :¼ ½1; 1Þ, but that it is not, uniformly continuous on B :¼ ð0; 1Þ., 3. Use the Nonuniform Continuity Criterion 5.4.2 to show that the following functions are not, uniformly continuous on the given sets., (a) f ðxÞ :¼ x2 ; A :¼ ½0; 1Þ ., (b) gðxÞ :¼ sinð1=xÞ; B :¼ ð0; 1Þ ., 4. Show that the function f ðxÞ :¼ 1=ð1 þ x2 Þ for x 2 R is uniformly continuous on R., 5. Show that if f and g are uniformly continuous on a subset A of R, then f þ g is uniformly, continuous on A., 6. Show that if f and g are uniformly continuous on A  R and if they are both bounded on A, then, their product fg is uniformly continuous on A., 7. If f ðxÞ :¼ x and gðxÞ :¼ sin x, show that both f and g are uniformly continuous on R, but that, their product fg is not uniformly continuous on R., 8. Prove that if f and g are each uniformly continuous on R, then the composite function f  g is, uniformly continuous on R., 9. If f is uniformly continuous on A  R, and j f ðxÞj  k > 0 for all x 2 A, show that 1=f is, uniformly continuous on A., 10. Prove that if f is uniformly continuous on a bounded subset A of R, then f is bounded on A., pffiffiffi, 11. If gðxÞ :¼ x for x 2 ½0; 1, show that there does not exist a constant K such that jgðxÞj , K jxj for all x 2 ½0; 1. Conclude that the uniformly continuous g is not a Lipschitz function, on [0, 1]., 12. Show that if f is continuous on [0, 1) and uniformly continuous on [a, 1) for some positive, constant a, then f is uniformly continuous on [0, 1)., 13. Let A  R and suppose that f : A ! R has the following property: for each e > 0 there exists a, function ge : A ! R such that ge is uniformly continuous on A and j f ðxÞ  ge ðxÞj < e for all, x 2 A. Prove that f is uniformly continuous on A., 14. A function f : R ! R is said to be periodic on R if there exists a number p > 0 such that, f ðx þ pÞ ¼ f ðxÞ for all x 2 R. Prove that a continuous periodic function on R is bounded and, uniformly continuous on R.

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C05, , 12/08/2010, , 14:20:5, , Page 149, , 5.5 CONTINUITY AND GAUGES, 15. Let, (a), (b), (c), , 149, , f and g be Lipschitz functions on A., Show that the sum f þ g is also a Lipschitz function on A., Show that if f and g are bounded on A, then the product fg is a Lipschitz function on A., Give an example of a Lipschitz function f on [0, 1) such that its square f 2 is not a Lipschitz, function., , 16. A function is called absolutely continuous on an interval I if for any e > 0 there exists a d > 0, such, P that for any pair-wise, P disjoint subintervals ½xk ; yk ; k ¼ 1; 2; . . . ; n, of I such that, j f ðxk Þ  f ðyk Þj < e. Show that if f satisfies a Lipschitz condition, jxk  yk j < d we have, on I, then f is absolutely continuous on I., , Section 5.5 Continuity and Gaugesy, We will now introduce some concepts that will be used later—especially in Chapters 7 and, 10 on integration theory. However, we wish to introduce the notion of a ‘‘gauge’’ now, because of its connection with the study of continuous functions. We first define the notion, of a tagged partition of an interval., 5.5.1 Definition A partition of an interval I :¼ ½a; b is a collection P ¼ fI 1 ; . . . ; I n g of, non-overlapping closed intervals whose union is [a, b]. We ordinarily denote the intervals, by I i :¼ ½xi1 ; xi , where, a ¼ x0 <    < xi1 < xi <    < xn ¼ b:, The points xi ði ¼ 0; . . . ; nÞ are called the partition points of P. If a point ti has been, chosen from each interval Ii, for i ¼ 1, . . . , n, then the points ti are called the tags and the, set of ordered pairs, , P ¼ fðI 1 ; t1 Þ; . . . ; ðI n ; tn Þg, is called a tagged partition of I. (The dot signifies that the partition is tagged.), The ‘‘fineness’’ of a partition P refers to the lengths of the subintervals in P. Instead of, requiring that all subintervals have length less than some specific quantity, it is often useful, to allow varying degrees of fineness for different subintervals Ii in P. This is accomplished, by the use of a ‘‘gauge,’’ which we now define., 5.5.2 Definition A gauge on I is a strictly positive function defined on I. If d is a gauge on, I, then a (tagged) partition P_ is said to be d-fine if, ð1Þ, , ti 2 I i  ½ti  dðti Þ; ti þ dðti Þ for, , i ¼ 1; . . . ; n :, , We note that the notion of d-fineness requires that the partition be tagged, so we do not, need to say ‘‘tagged partition’’ in this case., A gauge d on an interval I assigns an interval ½t  dðtÞ; t þ dðtÞ to each point t 2 I. The, d-fineness of a partition P_ requires that each subinterval Ii of P_ is contained in the interval, determined by the gauge d and the tag ti for that subinterval. This is indicated by the, inclusions in (1); see Figure 5.5.1. Note that the length of the subintervals is also controlled, by the gauge and the tags; the next lemma reflects that control., y, , This section can be omitted on a first reading of this chapter.

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C05, , 12/08/2010, , 14:20:5, , 150, , Page 150, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Figure 5.5.1, , Inclusion (1), , 5.5.3 Lemma If a partition P_ of I :¼ ½a; b is d-fine and x 2 I, then there exists a tag ti, in P_ such that jx  ti j  dðti Þ., Proof. If x 2 I, there exists a subinterval ½xi1 ; xi  from P_ that contains x. Since P_ is, d-fine, then, ti  dðti Þ  xi1  x  xi  ti þ dðti Þ;, , ð2Þ, , whence it follows that jx  ti j  dðti Þ., , Q.E.D., , In the theory of Riemann integration, we will use gauges d that are constant, functions to control the fineness of the partition; in the theory of the generalized, Riemann integral, the use of nonconstant gauges is essential. But nonconstant gauge, functions arise quite naturally in connection with continuous functions. For, let f : I !, R be continuous on I and let e > 0 be given. Then, for each point t 2 I there, exists de ðtÞ > 0 such that if jx  tj < de ðtÞ and x 2 I, then j f ðxÞ  f ðtÞj < e. Since de, is defined and is strictly positive on I, the function de is a gauge on I. Later in this, section, we will use the relations between gauges and continuity to give alternative, proofs of the fundamental properties of continuous functions discussed in Sections 5.3, and 5.4., 5.5.4 Examples (a) If d and g are gauges on I :¼ ½a; b and if 0 < dðxÞ  g ðxÞ for all, x 2 I, then every partition P_ that is d-fine is also g-fine. This follows immediately from the, inequalities, ti  gðti Þ  ti  dðti Þ and, , ti þ dðti Þ  ti þ gðti Þ, , which imply that, ti 2 ½ti  dðti Þ; ti þ dðti Þ  ½ti  gðti Þ; ti þ gðti Þ for, , i ¼ 1; . . . ; n :, , (b) If d1 and d2 are gauges on I :¼ ½a; b and if, dðxÞ :¼ minfd1 ðxÞ; d2 ðxÞg for all, , x2I;, , then d is also a gauge on I. Moreover, since dðxÞ  d1 ðxÞ, then every d-fine partition is d1 fine. Similarly, every d-fine partition is also d2 -fine., (c) Suppose that d is defined on I :¼ ½0; 1 by, (, dðxÞ :¼, , 1, 10, , if, , x ¼ 0;, , 1, 2x, , if, , 0 < x  1:, , , , Then d is a gauge on [0, 1]. If 0 < t  1, then ½t  dðtÞ; t þ dðtÞ ¼ 12 t; 32 t , which does not, contain the point 0. Thus, if P_ is a d-fine partition of I, then the only subinterval in P_ that, contains 0 must have the point 0 as its tag.

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C05, , 12/08/2010, , 14:20:7, , Page 151, , 5.5 CONTINUITY AND GAUGES, , 151, , (d) Let g be defined on I :¼ ½0; 1 by, 81, >, >, < 10, gðxÞ :¼ 12 x, >, >, :1, 2 ð1  xÞ, , if, , x ¼ 0 or x ¼ 1;, , if, , 0 < x  12 ;, , if, , 1, 2, , < x < 1:, , Then g is a gauge on I, and it is an exercise to show that the subintervals in any g-fine, partition that contain the points 0 or 1 must have these points as tags., &, , Existence of d-Fine Partitions, In view of the above examples, it is not obvious that an arbitrary gauge d admits a d-fine, partition. We now use the Supremum Property of R to establish the existence of d-fine, partitions. In the exercises, we will sketch a proof based on the Nested Intervals, Theorem 2.5.2., 5.5.5 Theorem If d is a gauge defined on the interval [a, b], then there exists a d-fine, partition of [a, b]., Proof. Let E denote the set of all points x 2 ½a; b such that there exists a d-fine partition, of the subinterval [a, x]. The set E is not empty, since the pair ([a, x], a) is a d-fine, partition of the interval [a, x] when x 2 ½a; a þ dðaÞ and x  b. Since E  ½a; b, the set, E is also bounded. Let u :¼ sup E so that a < u  b. We will show that u 2 E and that, u ¼ b., We claim that u 2 E. Since u  dðuÞ < u ¼ sup E, there exists v 2 E such that, u  dðuÞ < v < u. Let P_ 1 be a d-fine partition of [a, v] and let P_ 2 :¼ P_ 1 [ ð½v; u; uÞ., Then P_ 2 is a d-fine partition of [a, u], so that u 2 E., _ 1 is a d-fine partition of, If u < b, let w 2 ½a; b be such that u < w < u þ dðuÞ. If Q, _, _, _, [a, u], we let Q2 :¼ Q1 [ ð½u; w; uÞ. Then Q2 is a d-fine partition of [a, w], whence, w 2 E. But this contradicts the supposition that u is an upper bound of E. Therefore, u ¼ b., Q.E.D., , Some Applications, Following R. A. Gordon (see his Monthly article in the References), we will now show that, some of the major theorems in the two preceding sections can be proved by using gauges., Alternate Proof of Theorem 5.3.2: Boundedness Theorem. Since f is continuous on I,, then for each t 2 I there exists dðtÞ > 0 such that if x 2 I and jx  tj  dðtÞ, then, j f ðxÞ  f ðtÞj  1. Thus d is a gauge on I. Let fðI i ; ti Þgni¼1 be a d-fine partition of I and, let K :¼ maxfj f ðti Þj : i ¼ 1; . . . ; ng. By Lemma 5.5.3, given any x 2 I there exists i with, jx  ti j  dðti Þ, whence, j f ðxÞj  j f ðxÞ  f ðti Þj þ j f ðti Þj  1 þ K:, Since x 2 I is arbitrary, then f is bounded by 1 þ K on I., , Q.E.D.

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C05, , 12/08/2010, , 14:20:9, , 152, , Page 152, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Alternate Proof of Theorem 5.3.4: Maximum-Minimum Theorem. We will prove the, existence of x . Let M :¼ supff ðxÞ : x 2 I g and suppose that f(x) < M for all x 2 I. Since f, is continuous on I, for each t 2 I there exists dðtÞ > 0 such that if x 2 I and jx  tj  dðtÞ,, then f ðxÞ < 12 ðM þ f ðtÞÞ. Thus d is a gauge on I, and if fðI i ; ti Þgni¼1 is a d-fine partition of I,, we let, ~, M :¼ 12 maxfM þ f ðt1 Þ; . . . ; M þ f ðtn Þg:, By Lemma 5.5.3, given any x 2 I, there exists i with jx  ti j  dðti Þ, whence, ~, f ðxÞ < 12 ðM þ f ðti ÞÞ  M:, ~, Since x 2 I is arbitrary, then Mð< MÞ is an upper bound for f on I, contrary to the definition, of M as the supremum of f., Q.E.D., Alternate Proof of Theorem 5.3.5: Location of Roots Theorem. We assume that f ðtÞ 6¼ 0, for all t 2 I. Since f is continuous at t, Exercise 5.1.7 implies that there exists dðtÞ > 0 such, that if x 2 I and jx  tj  dðtÞ, then f (x) < 0 if f (t) < 0, and f (x) > 0 if f (t) > 0. Then d is, a gauge on I and we let fðI i ; ti Þgni¼1 be a d-fine partition. Note that for each i, either f (x) < 0, for all x 2 ½xi1 ; xi  or f (x) > 0 for all such x. Since f ðx0 Þ ¼ f ðaÞ < 0, this implies that, f (x1) < 0, which in turn implies that f (x2) < 0. Continuing in this way, we have, f ðbÞ ¼ f ðxn Þ < 0, contrary to the hypothesis that f (b) > 0., Q.E.D., Alternate Proof of Theorem 5.4.3: Uniform Continuity Theorem. Let e > 0 be given., Since f is continuous at t 2 I, there exists dðtÞ > 0 such that if x 2 I and jx  tj  2dðtÞ,, then j f ðxÞ  f ðtÞj  12 e. Thus d is a gauge on I. If fðI i ; ti Þgni¼1 is a d-fine partition of I,, let de :¼ minfdðt1 Þ; . . . ; dðtn Þg. Now suppose that x; u 2 I and jx  uj  de , and choose i, with jx  ti j  dðti Þ. Since, ju  ti j  ju  xj þ jx  ti j  de þ dðti Þ  2dðti Þ;, then it follows that, j f ðxÞ  f ðuÞj  j f ðxÞ  f ðti Þj þ j f ðti Þ  f ðuÞj  12 e þ 12 e ¼ e:, Therefore, f is uniformly continuous on I., , Q.E.D., , Exercises for Section 5.5, 1. Let d be the gauge on [0, 1] defined by dð0Þ: ¼ 14 and dðtÞ :¼ 12 t for t 2 ð0; 1.,  1, , , , , (a) Show that P_ 1 ¼, 0; 4 ; 0 ; 14 ; 12 ; 12 ; 12 ; 1 ; 34 is d-fine., ,  1, , , , (b) Show that P_ 2 ¼, 0; 4 ; 0 ; 14 ; 12 ; 12 ; 12 ; 1 ; 35 is not d-fine., 2. Suppose that d1 is the gauge defined by d1 ð0Þ :¼ 14 ; d1 ðtÞ :¼ 34 t for t 2 ð0; 1. Are the partitions, given in Exercise 1 d1 -fine? Note that dðtÞ  d1 ðtÞ for all t 2 ½0; 1., 1, 9, 3. Suppose that d2 is the gauge defined by d2 ð0Þ :¼ 10, and d2 ðtÞ :¼ 10, t for t 2 ð0; 1. Are the, partitions given in Exercise 1 d2 -fine?, , 4. Let g be the gauge in Example 5.5.4(d)., , , (a) If t 2 ð0; 12 show that ½t  gðtÞ; t þ gðtÞ ¼ 12 t; 32 t  ð0; 34., (b) If t 2 ð12 ; 1Þ show that ½t  gðtÞ; t þ gðtÞ  ð14 ; 1Þ., 5. Let a < c < b and let d be a gauge on [a, b]. If P_ 0 is a d-fine partition of [a, c] and if P_00 is a d-fine, partition of [c, b], show that P_ 0 [ P_00 is a d-fine partition of [a, b] having c as a partition point.

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C05, , 12/08/2010, , 14:20:10, , Page 153, , 5.6 MONOTONE AND INVERSE FUNCTIONS, , 153, , 6. Let a < c < b and let d0 and d00 be gauges on [a, c] and [c, b], respectively. If d is defined on, [a, b] by, 80, d ðtÞ, >, >, <, dðtÞ :¼ minfd0 ðcÞ; d00 ðcÞg, >, >, : 00, d ðtÞ, , if, , t 2 ½a; cÞ;, , if, , t ¼ c;, , if, , t 2 ðc; b;, , 00, then d is a gauge on [a, b]. Moreover, if P_ 0 is a d0 -fine partition of [a, c] and P_00 is a d -fine, 0, 00, partition of [c, b], then P_ [ P_ is a tagged partition of [a, b] having c as a partition point., Explain why P_ 0 [ P_00 may not be d-fine. Give an example., 00, , 7. Let d0 and d be as in the preceding exercise and let d be defined by, 8, min d0 ðtÞ; 12 ðc  tÞ, >, >, <, d ðtÞ :¼ minfd0 ðcÞ; d00 ðcÞg, >, >, :, min d00 ðtÞ; 12 ðt  cÞ, , if, , t 2 ½a; cÞ;, , if, , t ¼ c;, , if, , t 2 ðc; b:, , Show that d is a gauge on [a, b] and that every d -fine partition P_ of [a, b] having c as a partition, point gives rise to a d0 -fine partition P_ 0 of [a, c] and a d00 -fine partition P_ 00 of [c, b] such that, P_ ¼ P_ 0 [ P_00 ., 8. Let d be a gauge on I :¼ ½a; b and suppose that I does not have a d-fine partition., (a) Let c :¼ 12 ða þ bÞ. Show that at least one of the intervals [a, c] and [c, b] does not have a, d-fine partition., (b) Construct a nested sequence (In) of subintervals with the length of In equal to ðb  aÞ=2n, such that In does not have a d-fine partition., p, (c) Let j 2 \1, n¼1 I n and let,  p 2 N be such that ðb  aÞ=2 < dðjÞ. Show that I p  ½j  dðjÞ;, j þ dðjÞ, so the pair I p ; j is a d-fine partition of Ip., 9. Let I :¼ ½a; b and let f : I ! R be a (not necessarily continuous) function. We say that f is, ‘‘locally bounded’’ at c 2 I if there exists dðcÞ > 0 such that f is bounded on I \ ½c  dðcÞ;, c þ dðcÞ. Prove that if f is locally bounded at every point of I, then f is bounded on I., 10. Let I :¼ ½a; b and f : I ! R. We say that f is ‘‘locally increasing’’ at c 2 I if there exists dðcÞ >, 0 such that f is increasing on I \ ½c  dðcÞ; c þ dðcÞ. Prove that if f is locally increasing at every, point of I, then f is increasing on I., , Section 5.6 Monotone and Inverse Functions, Recall that if A  R, then a function f : A ! R is said to be increasing on A if whenever, x1 ; x2 2 A and x1  x2 , then f ðx1 Þ  f ðx2 Þ. The function f is said to be strictly increasing, on A if whenever x1 ;x2 2 A and x1 < x2 , then f ðx1 Þ < f ðx2 Þ. Similarly, g : A ! R is said, to be decreasing on A if whenever x1 ; x2 2 A and x1  x2 then gðx1 Þ  gðx2 Þ. The, function g is said to be strictly decreasing on A if whenever x1 ; x2 2 A and x1 < x2 then, gðx1 Þ > gðx2 Þ., If a function is either increasing or decreasing on A, we say that it is monotone on A., If f is either strictly increasing or strictly decreasing on A, we say that f is strictly, monotone on A., We note that if f : A ! R is increasing on A then g :¼ f is decreasing on A; similarly, if w : A ! R is decreasing on A then c :¼ w is increasing on A.

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C05, , 12/08/2010, , 14:20:11, , 154, , Page 154, , CHAPTER 5 CONTINUOUS FUNCTIONS, , In this section, we will be concerned with monotone functions that are defined on, an interval I  R. We will discuss increasing functions explicitly, but it is clear that, there are corresponding results for decreasing functions. These results can either be, obtained directly from the results for increasing functions or proved by similar, arguments., Monotone functions are not necessarily continuous. For example, if f ðxÞ :¼ 0 for, x 2 ½0; 1 and f ðxÞ :¼ 1 for x 2 ð1; 2, then f is increasing on [0, 2], but fails to be, continuous at x ¼ 1. However, the next result shows that a monotone function always has, both one-sided limits (see Definition 4.3.1) in R at every point that is not an endpoint of its, domain., 5.6.1 Theorem Let I  R be an interval and let f : I ! R be increasing on I. Suppose, that c 2 I is not an endpoint of I. Then, (i), (ii), , lim f ¼ supf f ðxÞ : x 2 I; x < cg,, , x!c, , lim f ¼ inf f f ðxÞ : x 2 I; x > cg., , x!cþ, , Proof. (i) First note that if x 2 I and x < c, then f ðxÞ  f ðcÞ. Hence the set, f f ðxÞ : x 2 I; x < cg, which is nonvoid since c is not an endpoint of I, is bounded above, by f (c). Thus the indicated supremum exists; we denote it by L. If e > 0 is given, then L  e, is not an upper bound of this set. Hence there exists ye 2 I; ye < c such that, L  e < f ðye Þ  L., Since f is increasing, we deduce that if de :¼ c  ye and if 0 < c  y < de , then ye <, y < c so that, L  e < f ðye Þ  f ðyÞ  L:, Therefore j f ðyÞ  Lj < e when 0 < c  y < de . Since e > 0 is arbitrary we infer that (i), holds., The proof of (ii) is similar., Q.E.D., The next result gives criteria for the continuity of an increasing function f at a point c, that is not an endpoint of the interval on which f is defined., 5.6.2 Corollary Let I  R be an interval and let f : I ! R be increasing on I. Suppose, that c 2 I is not an endpoint of I. Then the following statements are equivalent., (a) f is continuous at c., (b) lim f ¼ f ðcÞ ¼ lim f ., x!c, , x!cþ, , (c) supf f ðxÞ : x 2 I; x < cg ¼ f ðcÞ ¼ inf f f ðxÞ : x 2 I; x > cg., This follows easily from Theorems 4.3.3 and 5.6.1. We leave the details to the, reader., Let I be an interval and let f : I ! R be an increasing function. If a is the left endpoint, of I, it is an exercise to show that f is continuous at a if and only if, f ðaÞ ¼ inf f f ðxÞ : x 2 I; a < xg, or if and only if f ðaÞ ¼ lim f . Similar conditions apply at a right endpoint, and for, x!aþ, decreasing functions.

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C05, , 12/08/2010, , 14:20:12, , Page 155, , 5.6 MONOTONE AND INVERSE FUNCTIONS, , Figure 5.6.1, , 155, , The jump of f at c, , If f : I ! R is increasing on I and if c is not an endpoint of I, we define the jump of f, at c to be j f ðcÞ :¼ lim f  lim f . (See Figure 5.6.1.) It follows from Theorem 5.6.1, x!cþ, x!c, that, j f ðcÞ ¼ inf f f ðxÞ : x 2 I; x > cg  supf f ðxÞ : x 2 I; x < cg, for an increasing function. If the left endpoint a of I belongs to I, we define the jump of f at, a to be j f ðaÞ :¼ lim f  f ðaÞ. If the right endpoint b of I belongs to I, we define the jump, x!aþ, , of f at b to be j f ðbÞ :¼ f ðbÞ  lim f ., x!b, , 5.6.3 Theorem Let I  R be an interval and let f : I ! R be increasing on I. If c 2 I,, then f is continuous at c if and only if j f ðcÞ ¼ 0., Proof. If c is not an endpoint, this follows immediately from Corollary 5.6.2. If c 2 I is, the left endpoint of I, then f is continuous at c if and only if f ðcÞ ¼ lim f , which is, x!cþ, equivalent to j f ðcÞ ¼ 0. Similar remarks apply to the case of a right endpoint., Q.E.D., We now show that there can be at most a countable set of points at which a monotone, function is discontinuous., 5.6.4 Theorem Let I  R be an interval and let f : I ! R be monotone on I. Then the, set of points D  I at which f is discontinuous is a countable set., Proof. We shall suppose that f is increasing on I. It follows from Theorem 5.6.3 that, D ¼ x 2 I : j f ð xÞ 6¼ 0 . We shall consider the case that I :¼ ½a; b is a closed bounded, interval, leaving the case of an arbitrary interval to the reader., We first note that since f is increasing, then j f ðcÞ  0 for all c 2 I. Moreover, if, a  x1 <    xn  b, then (why?) we have, ð1Þ, , f ðaÞ  f ðaÞ þ j f ðx1 Þ þ    þ j f ðxn Þ  f ðbÞ;, , whence it follows that, j f ðx1 Þ þ    þ j f ðxn Þ  f ðbÞ  f ðaÞ:, (See Figure 5.6.2.) Consequently there can be at most k points in I ¼ ½a; b where, j f ðxÞ  ð f ðbÞ  f ðaÞÞ=k. We conclude that there is at most one point x 2 I where, j f ðxÞ ¼ f ðbÞ  f ðaÞ; there are at most two points in I where j f ðxÞ  ð f ðbÞ  f ðaÞÞ=2;, at most three points in I where j f ðxÞ  ð f ðbÞ  f ðaÞÞ=3, and so on. Therefore there is at

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C05, , 12/08/2010, , 14:20:12, , 156, , Page 156, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Figure 5.6.2, , j f ðx1 Þ þ    þ j f ðxn Þ  f ðbÞ  f ðaÞ, , most a countable set of points x where j f ðxÞ > 0. But since every point in D must be, included in this set, we deduce that D is a countable set., Q.E.D., Theorem 5.6.4 has some useful applications. For example, it was seen in Exercise, 5.2.12 that if h : R ! R satisfies the identity, ð2Þ, , hðx þ yÞ ¼ hðxÞ þ hðyÞ, , for all, , x; y 2 R ;, , and if h is continuous at a single point x0, then h is continuous at every point of R. Thus, if h, is a monotone function satisfying (2), then h must be continuous on R. [It follows from this, that h(x) ¼ Cx for all x 2 R, where C :¼ hð1Þ.], Inverse Functions, We shall now consider the existence of inverses for functions that are continuous on an, interval I  R. We recall (see Section 1.1) that a function f : I ! R has an inverse function, if and only if f is injective (¼ one-one); that is, x; y 2 I and x 6¼ y imply that f ðxÞ 6¼ f ðyÞ., We note that a strictly monotone function is injective and so has an inverse. In the next, theorem, we show that if f : I ! R is a strictly monotone continuous function, then f, has an inverse function g on J :¼ f ðIÞ that is strictly monotone and continuous on J. In, particular, if f is strictly increasing then so is g, and if f is strictly decreasing then so is g., 5.6.5 Continuous Inverse Theorem Let I  R be an interval and let f : I ! R be, strictly monotone and continuous on I. Then the function g inverse to f is strictly monotone, and continuous on J :¼ f ðIÞ., Proof. We consider the case that f is strictly increasing, leaving the case that f is strictly, decreasing to the reader., Since f is continuous and I is an interval, it follows from the Preservation of Intervals, Theorem 5.3.10 that J :¼ f ðIÞ is an interval. Moreover, since f is strictly increasing on I, it, is injective on I; therefore the function g : J ! R inverse to f exists. We claim that g is

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C05, , 12/08/2010, , 14:20:13, , Page 157, , 5.6 MONOTONE AND INVERSE FUNCTIONS, , Figure 5.6.3, , 157, , gðyÞ 6¼ x for y 2 J, , strictly increasing. Indeed, if y1 ; y2 2 J with y1 < y2 , then y1 ¼ f ðx1 Þ and y2 ¼ f ðx2 Þ for, some x1 ; x2 2 I. We must have x1 < x2 ; otherwise x1  x2 , which implies that, y1 ¼ f ðx1 Þ  f ðx2 Þ ¼ y2 , contrary to the hypothesis that y1 < y2 . Therefore we have, gðy1 Þ ¼ x1 < x2 ¼ gðy2 Þ. Since y1 and y2 are arbitrary elements of J with y1 < y2 , we, conclude that g is strictly increasing on J., It remains to show that g is continuous on J. However, this is a consequence of the fact, that gðJÞ ¼ I is an interval. Indeed, if g is discontinuous at a point c 2 J, then the jump of g, at c is nonzero so that lim g < lim g. If we choose any number x 6¼ gðcÞ satisfying, y!c, , y!cþ, , lim g < x < lim g, then x has the property that x 6¼ gðyÞ for any y 2 J. (See, x!cþ, Figure 5.6.3.) Hence x 2, = I, which contradicts the fact that I is an interval. Therefore, we conclude that g is continuous on J., Q.E.D., , x!c, , The nth Root Function, We will apply the Continuous Inverse Theorem 5.6.5 to the nth power function. We need to, distinguish two cases: (i) n even, and (ii) n odd., (i) n even. In order to obtain a function that is strictly monotone, we restrict our, attention to the interval I :¼ ½0; 1Þ. Thus, let f ðxÞ :¼ xn for x 2 I. (See Figure 5.6.4.) We, have seen (in Exercise 2.1.23) that if 0  x < y, then f ðxÞ ¼ xn < yn ¼ f ðyÞ; therefore f is, strictly increasing on I. Moreover, it follows from Example 5.2.3(a) that f is continuous on, I. Therefore, by the Preservation of Intervals Theorem 5.3.10, J :¼ f ðIÞ is an interval., We will show that J ¼ ½0; 1Þ. Let y  0 be arbitrary; by the Archimedean Property, there, exists k 2 N such that 0  y < k. Since, f ð0Þ ¼ 0  y < k  kn ¼ f ðkÞ;, it follows from Bolzano’s Intermediate Value Theorem 5.3.7 that y 2 J. Since y  0 is, arbitrary, we deduce that J ¼ ½0; 1Þ., We conclude from the Continuous Inverse Theorem 5.6.5 that the function g that is, inverse to f ðxÞ ¼ xn on I ¼ ½0; 1Þ is strictly increasing and continuous on J ¼ ½0; 1Þ. We, usually write, pffiffiffi, gðxÞ ¼ x1=n or gðxÞ ¼ n x

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C05, , 12/08/2010, , 14:20:14, , 158, , Page 158, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Figure 5.6.5 Graph of, gðxÞ ¼ x1=n ðx  0; n evenÞ, , Figure 5.6.4 Graph of, f ðxÞ ¼ xn ðx  0; n evenÞ, , pffiffiffi, for x  0 (n even), and call x1=n ¼ n x the nth root of x  0 (n even). The function g is, called the nth root function (n even). (See Figure 5.6.5.), Since g is inverse to f we have, gð f ðxÞÞ ¼ x, , and, , f ðgðxÞÞ ¼ x, , for all x 2 ½0; 1Þ:, , We can write these equations in the following form:, ðxn Þ1=n ¼ x, , and, , ðx1=n Þn ¼ x, , for all x 2 ½0; 1Þ and n even., (ii) n odd. In this case we let FðxÞ :¼ xn for all x 2 R; by 5.2.3(a), F is continuous, on R. We leave it to the reader to show that F is strictly increasing on R and that FðRÞ ¼ R., (See Figure 5.6.6.), It follows from the Continuous Inverse Theorem 5.6.5 that the function G that is, inverse to FðxÞ ¼ xn for x 2 R, is strictly increasing and continuous on R. We usually, write, pffiffiffi, GðxÞ ¼ x1=n or GðxÞ ¼ n x for x 2 R; n odd;, and call x1=n the nth root of x 2 R. The function G is called the nth root function (n odd)., (See Figure 5.6.7.) Here we have,  1=n n, ðxn Þ1=n ¼ x, and, x, ¼x, for all x 2 R and n odd., , Figure 5.6.6 Graph of, F(x) ¼ xn (x 2 R, n odd), , Figure 5.6.7 Graph of, G(x) ¼ x1=n (x 2 R, n odd)

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C05, , 12/08/2010, , 14:20:15, , Page 159, , 5.6 MONOTONE AND INVERSE FUNCTIONS, , 159, , Rational Powers, Now that the nth root functions have been defined for n 2 N, it is easy to define rational, powers., , 5.6.6 Definition (i) If m, n 2 N and x  0, we, define xm=n :¼ x1=n, , m, (ii) If m, n 2 N and x > 0, we define xm=n :¼ x1=n, ., , m, , ., , Hence we have defined xr when r is a rational number and x > 0. The graphs of x 7! xr, depend on whether r > 1, r ¼ 1, 0 < r < 1, r ¼ 0, or r < 0. (See Figure 5.6.8.) Since a, rational number r 2 Q can be written in the form r ¼ m=n with m 2 Z, n 2 N, in many, ways, it should be shown that Definition 5.6.6is not ambiguous., That is, if r ¼ m=n ¼ p=q, , m, p, with m, p 2 Z and n, q 2 N and if x > 0, then x1=n ¼ x1=q . We leave it as an exercise, to the reader to establish this relation., 5.6.7 Theorem If m 2 Z, n 2 N, and x > 0, then xm=n ¼ ðxm Þ1=n ., , m, Proof. If x > 0 and m, n 2 Z, then (xm)n ¼ xmn ¼ (xn)m. Now let y :¼ xm=n ¼ x1=n > 0, , m n  1=n n m, ¼ x, ¼ xm . Therefore it follows that y ¼ ðxm Þ1=n ., so that yn ¼ x1=n, Q.E.D., , The reader should also show, as an exercise, that if x > 0 and r, s 2 Q , then, xr xs ¼ xrþs ¼ xs xr, , Figure 5.6.8, , and, , ðxr Þs ¼ xrs ¼ ðxs Þr:, , Graphs of x ! xr, , ðx  0Þ

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C05, , 12/08/2010, , 14:20:16, , 160, , Page 160, , CHAPTER 5 CONTINUOUS FUNCTIONS, , Exercises for Section 5.6, 1. If I :¼ [a, b] is an interval and f : I ! R is an increasing function, then the point a [respectively,, b] is an absolute minimum [respectively, maximum] point for f on I. If f is strictly increasing,, then a is the only absolute minimum point for f on I., 2. If f and g are increasing functions on an interval I  R, show that f þ g is an increasing function, on I. If f is also strictly increasing on I, then f þ g is strictly increasing on I., 3. Show that both f (x) :¼ x and g(x) :¼ x  1 are strictly increasing on I :¼ [0, 1], but that their, product f g is not increasing on I., 4. Show that if f and g are positive increasing functions on an interval I, then their product fg is, increasing on I., 5. Show that if I :¼ [a, b] and f : I ! R is increasing on I, then f is continuous at a if and only if, f (a) ¼ inf{ f (x) : x 2 (a, b]}., 6. Let I  R be an interval and let f : I ! R be increasing on I. Suppose that c 2 I is not an, endpoint of I. Show that f is continuous at c if and only if there exists a sequence (xn) in I such, that xn < c for n ¼ 1, 3, 5, . . . ; xn > c for n ¼ 2, 4, 6, . . . ; and such that c ¼ lim(xn) and, f (c) ¼ lim ( f (xn))., 7. Let I  R be an interval and let f : I ! R be increasing on I. If c is not an endpoint of I,, show that the jump jf (c) of f at c is given by inf{ f (y)  f (x) : x < c < y, x, y 2 I}., 8. Let f, g be strictly increasing on an interval I  R and let f (x) > g(x) for all x 2 I. If, y 2 f ðI Þ \ gðI Þ, show that f 1 ðyÞ < g1 ðyÞ. [Hint: First interpret this statement geometrically.], 9. Let I :¼ [0, 1] and let f : I ! R be defined by f (x) :¼ x for x rational, and f (x) :¼ 1  x for x, irrational. Show that f is injective on I and that f ( f (x)) ¼ x for all x 2 I. (Hence f is its own, inverse function!) Show that f is continuous only at the point x ¼ 12., 10. Let I :¼ [a, b] and let f : I ! R be continuous on I. If f has an absolute maximum [respectively,, minimum] at an interior point c of I, show that f is not injective on I., 11. Let f (x) :¼ x for x 2 [0, 1], and f (x) :¼ 1 þ x for x 2 (1, 2]. Show that f and f 1 are strictly, increasing. Are f and f 1 continuous at every point?, 12. Let f : ½0; 1 ! R be a continuous function that does not take on any of its values twice and with, f (0) < f (1). Show that f is strictly increasing on [0, 1]., 13. Let h : ½0; 1 ! R be a function that takes on each of its values exactly twice. Show that h, cannot be continuous at every point. [Hint: If c1 < c2 are the points where h attains its, supremum, show that c1 ¼ 0, c2 ¼ 1. Now examine the points where h attains its infimum.], , , m, p, 14. Let x 2 R, x > 0. Show that if m, p 2 Z, n, q 2 N, and mq ¼ np, then x1=n ¼ x1=q ., 15. If x 2 R, x > 0, and if r, s 2 Q , show that x r x s ¼ x rþs ¼ x s x r and ðx r Þs ¼ x rs ¼ ðx s Þr .

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C06, , 12/09/2010, , 14:53:55, , Page 161, , CHAPTER 6, , DIFFERENTIATION, , Prior to the seventeenth century, a curve was generally described as a locus of points, satisfying some geometric condition, and tangent lines were obtained through geometric, construction. This viewpoint changed dramatically with the creation of analytic geometry in, the 1630s by Ren
e Descartes (1596–1650) and Pierre de Fermat (1601–1665). In this new, setting geometric problems were recast in terms of algebraic expressions, and new classes of, curves were defined by algebraic rather than geometric conditions. The concept of derivative, evolved in this new context. The problem of finding tangent lines and the seemingly unrelated, problem of finding maximum or minimum values were first seen to have a connection by, Fermat in the 1630s. And the relation between tangent lines to curves and the velocity of a, moving particle was discovered in the late 1660s by Isaac Newton. Newton’s theory of, ‘‘fluxions,’’ which was based on an intuitive idea of limit, would be familiar to any modern, student of differential calculus once some changes in terminology and notation were made., But the vital observation, made by Newton and, independently, by Gottfried Leibniz in the, 1680s, was that areas under curves could be calculated by reversing the differentiation, process. This exciting technique, one that solved previously difficult area problems with ease,, sparked enormous interest among the mathematicians of the era and led to a coherent theory, that became known as the differential and integral calculus., , Isaac Newton, Isaac Newton (1642–1727) was born in Woolsthorpe, in Lincolnshire,, England, on Christmas Day; his father, a farmer, had died three months, earlier. His mother remarried when he was three years old and he was sent, to live with his grandmother. He returned to his mother at age eleven, only, to be sent to boarding school in Grantham the next year. Fortunately, a, perceptive teacher noticed his mathematical talent and, in 1661, Newton, entered Trinity College at Cambridge University, where he studied with, Isaac Barrow., When the bubonic plague struck in 1665–1666, leaving dead nearly, 70,000 persons in London, the university closed and Newton spent two, years back in Woolsthorpe. It was during this period that he formulated his, basic ideas concerning optics, gravitation, and his method of ‘‘fluxions,’’ later called ‘‘calculus.’’, He returned to Cambridge in 1667 and was appointed Lucasian Professor in 1669. His theories of, universal gravitation and planetary motion were published to world acclaim in 1687 under the title, Philosophies Naturalis Principia Mathematica. However, he neglected to publish his method of, inverse tangents for finding areas and other work in calculus, and this led to a controversy over, priority with Leibniz., Following an illness, he retired from Cambridge University and in 1696 was appointed, Warden of the British mint. However, he maintained contact with advances in science and, mathematics and served as President of the Royal Society from 1703 until his death in 1727. At, his funeral, Newton was eulogized as ‘‘the greatest genius that ever existed.’’ His place of burial in, Westminster Abbey is a popular tourist site., , 161

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C06, , 12/09/2010, , 14:53:55, , 162, , Page 162, , CHAPTER 6 DIFFERENTIATION, , In this chapter we will develop the theory of differentiation. Integration theory,, including the fundamental theorem that relates differentiation and integration, will be the, subject of the next chapter. We will assume that the reader is already familiar with the, geometrical and physical interpretations of the derivative of a function as described in, introductory calculus courses. Consequently, we will concentrate on the mathematical, aspects of the derivative and not go into its applications in geometry, physics, economics,, and so on., The first section is devoted to a presentation of the basic results concerning the, differentiation of functions. In Section 6.2 we discuss the fundamental Mean Value, Theorem and some of its applications. In Section 6.3 the important L’Hospital Rules, are presented for the calculation of certain types of ‘‘indeterminate’’ limits., In Section 6.4 we give a brief discussion of Taylor’s Theorem and a few of its, applications—for example, to convex functions and to Newton’s Method for the location, of roots., , Section 6.1 The Derivative, In this section we will present some of the elementary properties of the derivative. We, begin with the definition of the derivative of a function., 6.1.1 Definition Let I
R be an interval, let f : I ! R, and let c 2 I. We say that a real, number L is the derivative of f at c if given any e > 0 there exists dðeÞ > 0 such that if x 2 I, satisfies 0 < jx  cj < dðeÞ, then,
,
,
f ðxÞ  f ðcÞ,
,
, ð1Þ,  L

< e:,
xc, In this case we say that f is differentiable at c, and we write f 0 ðcÞ for L., In other words, the derivative of f at c is given by the limit, f 0 ðcÞ ¼ lim, , ð2Þ, , x!c, , f ðxÞ  f ðcÞ, xc, , provided this limit exists. (We allow the possibility that c may be the endpoint of the, interval.), Note It is possible to define the derivative of a function having a domain more general, than an interval (since the point c need only be an element of the domain and also a cluster, point of the domain) but the significance of the concept is most naturally apparent for, functions defined on intervals. Consequently we shall limit our attention to such functions., Whenever the derivative of f : I ! R exists at a point c 2 I, its value is denoted by, f 0 ðcÞ. In this way we obtain a function f 0 whose domain is a subset of the domain of f. In, working with the function f 0 , it is convenient to regard it also as a function of x. For, example, if f ðxÞ :¼ x2 for x 2 R, then at any c in R we have, f ðxÞ  f ðcÞ, x2  c2, ¼ lim, ¼ lim ðx þ cÞ ¼ 2c:, x!c, x!c x  c, x!c, xc, , f 0 ðcÞ ¼ lim, , Thus, in this case, the function f 0 is defined on all of R and f 0 ðxÞ ¼ 2x for x 2 R .

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C06, , 12/09/2010, , 14:53:55, , Page 163, , 6.1 THE DERIVATIVE, , 163, , We now show that continuity of f at a point c is a necessary (but not sufficient), condition for the existence of the derivative at c., 6.1.2 Theorem If f : I ! R has a derivative at c 2 I, then f is continuous at c., Proof. For all x 2 I; x 6¼ c, we have, f ðxÞ  f ðcÞ ¼, , , , f ðxÞ  f ðcÞ, ðx  cÞ:, xc, , Since f 0 ðcÞ exists, we may apply Theorem 4.2.4 concerning the limit of a product to, conclude that, , , , f ðxÞ  f ðcÞ , lim ð f ðxÞ  f ðcÞÞ ¼ lim, lim ðx  cÞ, x!c, x!c, x!c, xc, ¼ f 0 ðcÞ  0 ¼ 0:, Therefore, lim f ðxÞ ¼ f ðcÞ so that f is continuous at c., x!c, , Q.E.D., , The continuity of f : I ! R at a point does not assure the existence of the derivative at, that point. For example, if f ðxÞ :¼ jxj for x 2 R, then for x 6¼ 0 we have, ð f ðxÞ  f ð0ÞÞ=ðx  0Þ ¼ jxj=x, which is equal to 1 if x > 0, and equal to 1 if x < 0., Thus the limit at 0 does not exist [see Example 4.1.10(b)], and therefore the function is not, differentiable at 0. Hence, continuity at a point c is not a sufficient condition for the, derivative to exist at c., Remark By taking simple algebraic combinations of functions of the form x 7! jx  cj,, it is not difficult to construct continuous functions that do not have a derivative at a finite (or, even a countable) number of points. In 1872, Karl Weierstrass astounded the mathematical, world by giving an example of a function that is continuous at every point but whose, derivative does not exist anywhere. Such a function defied geometric intuition about, curves and tangent lines, and consequently spurred much deeper investigations into the, concepts of real analysis. It can be shown that the function f defined by the series, f ðxÞ :¼, , 1, X, 1, n, n cos ð3 xÞ, 2, n¼0, , has the stated property. A very interesting historical discussion of this and other examples, of continuous, nondifferentiable functions is given in Kline, pp. 955–966, and also in, Hawkins, pp. 44–46. A detailed proof for a slightly different example can be found in, Appendix E., There are a number of basic properties of the derivative that are very useful in the, calculation of the derivatives of various combinations of functions. We now provide the, justification of some of these properties, which will be familiar to the reader from earlier, courses., 6.1.3 Theorem Let I
R be an interval, let c 2 I, and let f : I ! R and g : I ! R be, functions that are differentiable at c. Then:, (a) If a 2 R, then the function a f is differentiable at c, and, ð3Þ, , ða f Þ0 ðcÞ ¼ a f 0 ðcÞ:

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C06, , 12/09/2010, , 14:53:56, , Page 165, , 6.1 THE DERIVATIVE, , 165, , (b) The function f 1 f 2    f n is differentiable at c, and, ð8Þ, , ð f 1 f 2    f n Þ0 ðcÞ ¼ f 01 ðcÞ f 2 ðcÞ    f n ðcÞ þ f 1 ðcÞ f 02 ðcÞ    f n ðcÞ, þ    þ f 1 ðcÞ f 2 ðcÞ    f 0n ðcÞ:, , An important special case of the extended product rule (8) occurs if the functions are, equal, that is, f 1 ¼ f 2 ¼    ¼ f n ¼ f . Then (8) becomes, ð f n Þ0 ðcÞ ¼ nð f ðcÞÞn1 f 0 ðcÞ:, , ð9Þ, , In particular, if we take f ðxÞ :¼ x, then we find the derivative of gðxÞ :¼ xn to be, g0 ðxÞ ¼ nxn1; n 2 N. The formula is extended to include negative integers by applying, the Quotient Rule 6.1.3(d)., Notation If I
R is an interval and f : I ! R, we have introduced the notation f 0 to, denote the function whose domain is a subset of I and whose value at a point c is the derivative, f 0 ðcÞ of f at c. There are other notations that are sometimes used for f 0; for example, one, sometimes writes Df for f 0. Thus one can write formulas (4) and (5) in the form:, Dð f þ gÞ ¼ Df þ Dg;, , Dð f gÞ ¼ ðDf Þ  g þ f  ðDgÞ:, , When x is the ‘‘independent variable,’’ it is common practice in elementary courses to write, df =dx for f 0. Thus formula (5) is sometimes written in the form, , , , , d, df, dg, ð f ðxÞgðxÞÞ ¼, ðxÞ gðxÞ þ f ðxÞ, ðxÞ :, dx, dx, dx, This last notation, due to Leibniz, has certain advantages. However, it also has certain, disadvantages and must be used with some care., The Chain Rule, We now turn to the theorem on the differentiation of composite functions known as the, ‘‘Chain Rule.’’ It provides a formula for finding the derivative of a composite function, g  f in terms of the derivatives of g and f., We first establish the following theorem concerning the derivative of a function at a, point that gives us a very nice method for proving the Chain Rule. It will also be used to, derive the formula for differentiating inverse functions., 6.1.5 Carath
eodory’s Theorem Let f be defined on an interval I containing the point c., Then f is differentiable at c if and only if there exists a function w on I that is continuous at, c and satisfies, ð10Þ, , f ðxÞ  f ðcÞ ¼ wðxÞðx  cÞ, , f or, , x 2 I:, , In this case, we have wðcÞ ¼ f 0 ðcÞ., Proof. ð)Þ If f 0 ðcÞ exists, we can define w by, 8, < f ðxÞ  f ðcÞ, for, wðxÞ :¼, xc, : 0, f ð cÞ, for, , x 6¼ c; x 2 I;, x ¼ c:, , The continuity of w follows from the fact that lim wðxÞ ¼ f 0 ðcÞ. If x ¼ c, then both sides of, x!c, , (10) equal 0, while if x 6¼ c, then multiplication of wðxÞ by x  c gives (10) for all other x 2 I.

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C06, , 12/09/2010, , 14:53:57, , Page 167, , 6.1 THE DERIVATIVE, , 167, , 6.1.7 Examples (a) If f : I ! R is differentiable on I and gðyÞ :¼ yn for y 2 R and, n 2 N, then since g0 ðyÞ ¼ nyn1 , it follows from the Chain Rule 6.1.6 that, ðg  f Þ0 ðxÞ ¼ g0 ð f ðxÞÞ  f 0 ðxÞ, , x 2 I:, , for, , Therefore we have ð f n Þ0 ðxÞ ¼ nð f ðxÞÞn1 f 0 ðxÞ for all x 2 I as was seen in (9)., (b) Suppose that f : I ! R is differentiable on I and that f ðxÞ 6¼ 0 and f 0 ðxÞ 6¼ 0 for x 2 I., If hðyÞ :¼ 1=y for y 6¼ 0, then it is an exercise to show that h0 ðyÞ ¼ 1=y2 for y 2 R; y 6¼ 0., Therefore we have,  0, 1, f 0 ðxÞ, ðxÞ ¼ ðh  f Þ0 ðxÞ ¼ h0 ð f ðxÞÞ f 0 ðxÞ ¼ , for x 2 I:, f, ð f ðxÞÞ2, (c) The absolute value function gðxÞ :¼ jxj is differentiable at all x 6¼ 0 and has derivative, g0 ðxÞ ¼ sgnðxÞ for x 6¼ 0. (The signum function is defined in Example 4.1.10(b).) Though, sgn is defined everywhere, it is not equal to g0 at x ¼ 0 since g0 ð0Þ does not exist., Now if f is a differentiable function, then the Chain Rule implies that the function, g  f ¼ j f j is also differentiable at all points x where f ðxÞ 6¼ 0, and its derivative is, given by, j f j0 ðxÞ ¼ sgnð f ðxÞÞ  f 0 ðxÞ ¼, , f 0 ðxÞ, 0, , f ðxÞ, , if, , f ðxÞ > 0;, , if, , f ðxÞ < 0:, , If f is differentiable at a point c with f ðcÞ ¼ 0, then it is an exercise to show that j f j is, differentiable at c if and only if f 0 ðcÞ ¼ 0. (See Exercise 7.), 2, For example,,
2,
if f ðxÞ :¼ x 0  1 for x 22R, then the derivative of its absolute value,
,
, j f jðxÞ ¼ x  1 is equal to j f j ðxÞ ¼ sgnðx  1Þ  ð2xÞ for x 6¼ 1; 1. See Figure 6.1.1, for a graph of j f j., , Figure 6.1.1, ,
,
, The function j f jðxÞ ¼
x2  1
., , (d) It will be proved later that if SðxÞ :¼ sin x and CðxÞ :¼ cos x for all x 2 R, then, S0 ðxÞ ¼ cos x ¼ C ðxÞ, , and C 0 ðxÞ ¼ sin x ¼ SðxÞ, , for all x 2 R. If we use these facts together with the definitions, tan x :¼, , sin x, 1, ; sec x :¼, ;, cos x, cos x

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C06, , 12/09/2010, , 14:53:58, , 168, , Page 168, , CHAPTER 6 DIFFERENTIATION, , for x 6¼ ð2k þ 1Þp=2; k 2 Z , and apply the Quotient Rule 6.1.3(d), we obtain, D tan x ¼, , ðcos xÞðcos xÞ  ðsin xÞðsin xÞ, 2, , ¼ ðsec xÞ2 ;, , ðcos xÞ, 0  1ðsin xÞ, sin x, ¼, ¼ ðsec xÞðtan xÞ, D sec x ¼, 2, ðcos xÞ, ðcos xÞ2, , for x 6¼ ð2k þ 1Þp=2; k 2 Z ., Similarly, since, cot x :¼, , cos x, ;, sin x, , csc x :¼, , 1, sin x, , for x 6¼ kp; k 2 Z , then we obtain, D cot x ¼ ðcsc xÞ2, , D csc x ¼ ðcsc xÞðcot xÞ, , and, , for x 6¼ kp; k 2 Z ., (e) Suppose that f is defined by, f ðxÞ :¼, , x2 sin ð1=xÞ, 0, , for, for, , x 6¼ 0;, x ¼ 0:, , If we use the fact that D sin x ¼ cos x for all x 2 R and apply the Product Rule 6.1.3(c) and, the Chain Rule 6.1.6, we obtain (why?), f 0 ðxÞ ¼ 2x sin ð1=xÞ  cos ð1=xÞ for, , x 6¼ 0:, , If x ¼ 0, none of the calculational rules may be applied. (Why?) Consequently, the, derivative of f at x ¼ 0 must be found by applying the definition of derivative. We find that, f ðxÞ  f ð0Þ, x2 sin ð1=xÞ, ¼ lim, ¼ lim x sin ð1=xÞ ¼ 0:, x!0, x!0, x!0, x0, x, , f 0 ð0Þ ¼ lim, , Hence, the derivative f 0 of f exists at all x 2 R. However, the function f 0 does not have a, limit at x ¼ 0 (why?), and consequently f 0 is discontinuous at x ¼ 0. Thus, a function f that, &, is differentiable at every point of R need not have a continuous derivative f 0., Inverse Functions, We will now relate the derivative of a function to the derivative of its inverse function, when, this inverse function exists. We will limit our attention to a continuous strictly monotone, function and use the Continuous Inverse Theorem 5.6.5 to ensure the existence of a, continuous inverse function., If f is a continuous strictly monotone function on an interval I, then its inverse function, g ¼ f 1 is defined on the interval J :¼ f ðI Þ and satisfies the relation, g f ðxÞ ¼ x, , for, , x 2 I:, , If c 2 I and d :¼ f ðcÞ, and if we knew that both f 0 ðcÞ and g0 ðd Þ exist, then we could, differentiate both sides of the equation and apply the Chain Rule to the left side to get, g0 f ðcÞ  f 0 ðcÞ ¼ 1. Thus, if f 0 ðcÞ 6¼ 0, we would obtain, g0 ð d Þ ¼, , 1, :, f 0 ðcÞ

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C06, , 12/09/2010, , 14:53:58, , Page 169, , 6.1 THE DERIVATIVE, , 169, , However, it is necessary to deduce the differentiability of the inverse function g from the, assumed differentiability of f before such a calculation can be performed. This is nicely, accomplished by using Carath
eodory’s Theorem., 6.1.8 Theorem Let I be an interval in R and let f : I ! R be strictly monotone and, continuous on I. Let J :¼ f ðI Þ and let g : J ! R be the strictly monotone and continuous, function inverse to f. If f is differentiable at c 2 I and f 0 ðcÞ 6¼ 0, then g is differentiable at, d :¼ f ðcÞ and, g0 ð d Þ ¼, , ð12Þ, , 1, 1, ¼, :, f 0 ðcÞ f 0 ðgðd ÞÞ, , Proof. Given c 2 R, we obtain from Carath
eodory’s Theorem 6.1.5 a function w on I with, properties that w is continuous at c; f ðxÞ  f ðcÞ ¼ wðxÞðx  cÞ for x 2 I, and wðcÞ ¼ f 0 ðcÞ., Since wðcÞ 6¼ 0 by hypothesis, there exists a neighborhood V :¼ ðc  d; c þ dÞ such that, wðxÞ 6¼ 0 for all x 2 V \ I. (See Theorem 4.2.9.) If U :¼ f ðV \ I Þ, then the inverse, function g satisfies f ðgðyÞÞ ¼ y for all y 2 U, so that, y  d ¼ f ðgðyÞÞ  f ðcÞ ¼ wðgðyÞÞ  ðgðyÞ  gðd ÞÞ:, Since wðgðyÞÞ 6¼ 0 for y 2 U, we can divide to get, gðyÞ  gðd Þ ¼, , 1,  ðy  d Þ:, w ð gð y Þ Þ, , Since the function 1=ðw  gÞ is continuous at d, we apply Theorem 6.1.5 to conclude that, g0 ðd Þ exists and g0 ðd Þ ¼ 1=wðgðd ÞÞ ¼ 1=wðcÞ ¼ 1=f 0 ðcÞ:, Q.E.D., Note The hypothesis, made in Theorem 6.1.8, that f 0 ðcÞ 6¼ 0 is essential. In fact, if, f 0 ðcÞ ¼ 0, then the inverse function g is never differentiable at d ¼ f ðcÞ, since the assumed, existence of g0 ðd Þ would lead to 1 ¼ f 0 ðcÞg0 ðd Þ ¼ 0, which is impossible. The function, f ðxÞ :¼ x3 with c ¼ 0 is such an example., 6.1.9 Theorem Let I be an interval and let f : I ! R be strictly monotone on I. Let, J :¼ f ðI Þ and let g : J ! R be the function inverse to f. If f is differentiable on I and, f 0 ðxÞ 6¼ 0 for x 2 I, then g is differentiable on J and, g0 ¼, , ð13Þ, , 1, :, f g, 0, , Proof. If f is differentiable on I, then Theorem 6.1.2 implies that f is continuous on I, and, by the Continuous Inverse Theorem 5.6.5, the inverse function g is continuous on J., Equation (13) now follows from Theorem 6.1.8., Q.E.D., Remark If f and g are the functions of Theorem 6.1.9, and if x 2 I and y 2 J are related, by y ¼ f ðxÞ and x ¼ gðyÞ, then equation (13) can be written in the form, g0 ð y Þ ¼, , 1, ; y 2 J;, ð f 0  gÞ ð y Þ, , or, , ðg0  f ÞðxÞ ¼, , 1, ; x 2 I:, f 0 ðxÞ, , It can also be written in the form g0 ðyÞ ¼ 1=f 0 ðxÞ, provided that it is kept in mind that x and, y are related by y ¼ f ðxÞ and x ¼ gðyÞ.

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C06, , 12/09/2010, , 14:53:59, , 170, , Page 170, , CHAPTER 6 DIFFERENTIATION, , 6.1.10 Examples (a) The function f : R ! R defined by f ðxÞ :¼ x5 þ 4x þ 3 is continuous and strictly monotone increasing (since it is the sum of two strictly increasing, functions). Moreover, f 0 ðxÞ ¼ 5x4 þ 4 is never zero. Therefore, by Theorem 6.1.8, the, inverse function g ¼ f 1 is differentiable at every point. If we take c ¼ 1, then since, f ð1Þ ¼ 8, we obtain g0 ð8Þ ¼ g0 ð f ð1ÞÞ ¼ 1=f 0 ð1Þ ¼ 1=9., (b) Let n 2 N be even, let I :¼ ½0; 1Þ, and let f ðxÞ :¼ xn for x 2 I. It was seen at the end, of Section 5.6 that f is strictly increasing and continuous on I, so that its inverse function, gðyÞ :¼ y1=n for y 2 J :¼ ½0; 1Þ is also strictly increasing and continuous on J. Moreover,, we have f 0 ðxÞ ¼ nx n1 for all x 2 I. Hence it follows that if y > 0, then g0 ðyÞ exists and, g0 ðyÞ ¼, , 1, 1, 1, ¼, ¼, :, f 0 ðgðyÞÞ nðgðyÞÞn1 nyðn1Þ=n, , Hence we deduce that, 1, g0 ðyÞ ¼ yð1=nÞ1 for y > 0:, n, However, g is not differentiable at 0. (For a graph of f and g, see Figures 5.6.4 and 5.6.5.), (c) Let n 2 N, n 6¼ 1, be odd, let F ðxÞ :¼ xn for x 2 R, and let GðyÞ :¼ y1=n be its inverse, function defined for all y 2 R. As in part (b) we find that G is differentiable for y 6¼ 0 and, that G0 ðyÞ ¼ ð1=nÞyð1=nÞ1 for y 6¼ 0. However, G is not differentiable at 0, even though G is, differentiable for all y 6¼ 0. (For a graph of F and G, see Figures 5.6.6 and 5.6.7.), (d) Let r :¼ m=n be a positive rational number, let I :¼ ½0; 1Þ, and let RðxÞ :¼ xr for, x 2 I. (Recall Definition 5.6.6.) Then R is the composition of the functions f ðxÞ :¼ xm and, gðxÞ :¼ x1=n ; x 2 I. That is, RðxÞ ¼ f ðgðxÞÞ for x 2 I. If we apply the Chain Rule 6.1.6, and the results of (b) [or (c), depending on whether n is even or odd], then we obtain, m1 1 ð1=nÞ1, R0 ðxÞ ¼ f 0 ðgðxÞÞg0 ðxÞ ¼ m x1=n,  x, n, m ðm=nÞ1, r1, ¼ x, ¼ rx, n, for all x > 0. If r > 1, then it is an exercise to show that the derivative also exists at x ¼ 0, and R0 ð0Þ ¼ 0. (For a graph of R see Figure 5.6.8.), (e) The sine function is strictly increasing on the interval I :¼ ½p=2; p=2; therefore its, inverse function, which we will denote by Arcsin, exists on J :¼ ½1; 1. That is, if x 2, ½p=2; p=2 and y 2 ½1; 1 then y ¼ sin x if and only if Arcsin y ¼ x. It was asserted, (without proof) in Example 6.1.7(d) that sin is differentiable on I and that D sin x ¼ cos x, for x 2 I. Since cos x 6¼ 0 for x in ðp=2; p=2Þ it follows from Theorem 6.1.8 that, , 1, 1, ¼, D sin x cos x, 1, 1, ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi, 1  y2, 1  ðsin xÞ2, , D Arcsin y ¼, , for all y 2 ð1; 1Þ. The derivative of Arcsin does not exist at the points 1 and 1., Exercises for Section 6.1, 1. Use the definition to find the derivative of each of the following functions:, (a) f ðxÞ :¼ xp3ffiffiffifor x 2 R;, (b) gðxÞ :¼ 1=x, pfor, ffiffiffi x 2 R; x 6¼ 0;, (c) hðxÞ :¼ x for x > 0;, (d) kðxÞ :¼ 1= x for x > 0:, , &

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C06, , 12/09/2010, , 14:54:0, , Page 171, , 6.1 THE DERIVATIVE, , 171, , 2. Show that f ðxÞ :¼ x1=3 ; x 2 R, is not differentiable at x ¼ 0., 3. Prove Theorem 6.1.3(a), (b)., 4. Let f : R ! R be defined by f ðxÞ :¼ x2 for x rational, f ðxÞ :¼ 0 for x irrational. Show that f is, differentiable at x ¼ 0, and find f 0 ð0Þ., 5. Differentiate and simplify:, x, (a) f ðxÞ :¼, ;, 1 þ x2, m, (c) hðxÞ :¼ sin xk for m; k 2 N;, , pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, 5  2x þ x2 ;, , (b), , gðxÞ :¼, , (d), , kðxÞ :¼ tan ðx2 Þ for jxj <, , pffiffiffiffiffiffiffiffi, p=2:, , 6. Let n 2 N and let f : R ! R be defined by f ðxÞ :¼ xn for x  0 and f ðxÞ :¼ 0 for x < 0. For, which values of n is f 0 continuous at 0? For which values of n is f 0 differentiable at 0?, 7. Suppose that f : R ! R is differentiable at c and that f ðcÞ ¼ 0. Show that gðxÞ :¼ j f ðxÞj is, differentiable at c if and only if f 0 ðcÞ ¼ 0., 8. Determine where each of the following functions from R to R is differentiable and find the, derivative:, (a) f ðxÞ :¼ jxj þ jx þ 1j;, (b) gðxÞ :¼ 2x þ jxj;, (c), , hðxÞ :¼ xjxj;, , (d), , kðxÞ :¼ jsin xj:, , 9. Prove that if f : R ! R is an even function [that is, f ðxÞ ¼ f ðxÞ for all x 2 R] and has a, derivative at every point, then the derivative f 0 is an odd function [that is, f 0 ðxÞ ¼ f 0 ðxÞ for, all x 2 R]. Also prove that if g : R ! R is a differentiable odd function, then g0 is an even, function., 10. Let g : R ! R be defined by gðxÞ :¼ x2 sin ð1=x2 Þ for x 6¼ 0, and gð0Þ :¼ 0. Show that g is, differentiable for all x 2 R. Also show that the derivative g0 is not bounded on the interval, ½1; 1., 11. Assume that there exists a function L : ð0; 1Þ ! R such that L0 ðxÞ ¼ 1=x for x > 0. Calculate, the derivatives of the following functions:, 3, (a) f ðxÞ :¼ Lð2x þ 3Þ for x > 0;, (b) gðxÞ :¼ ðLðx2 ÞÞ for x > 0;, (c), , hðxÞ :¼ LðaxÞ for a > 0; x > 0;, , (d), , kðxÞ :¼ LðLðxÞÞ when LðxÞ > 0; x > 0:, , 12. If r > 0 is a rational number, let f : R ! R be defined by f ðxÞ :¼ xr sin ð1=xÞ for x 6¼ 0, and, f ð0Þ :¼ 0. Determine those values of r for which f 0 ð0Þ exists., 13. If f : R ! R is differentiable at c 2 R, show that, f 0 ðcÞ ¼ limðnf f ðc þ 1=nÞ  f ðcÞgÞ:, However, show by example that the existence of the limit of this sequence does not imply the, existence of f 0 ðcÞ., 14. Given that the function hðxÞ :¼ x3 þ 2x þ 1 for x 2 R has an inverse h1 on R, find the value of, 0, h1 ðyÞ at the points corresponding to x ¼ 0; 1; 1., 15. Given that the restriction of the cosine function cos to I :¼ ½0; p is strictly decreasing and that, cos 0 ¼ 1; cos p ¼ 1, let J :¼ ½1; 1, and let Arccos: J ! R be the function inverse to the, restriction of cos to I. Show that Arccos is differentiable on ð1; 1Þ and D Acrccos y ¼, 1=2, ð1Þ=ð1  y2 Þ for y 2 ð1; 1Þ. Show that Arccos is not differentiable at 1 and 1., 16. Given that the restriction of the tangent function tan to I :¼ ðp=2; p=2Þ is strictly increasing, and that tan ðI Þ ¼ R, let Arctan: R ! R be the function inverse to the restriction of tan to I., Show that Arctan is differentiable on R and that D ArctanðyÞ ¼ ð1 þ y2 Þ1 for y 2 R., 17. Let f : I ! R be differentiable at c 2 I. Establish the Straddle Lemma: Given e > 0 there, exists dðeÞ > 0 such that if u, v 2 I satisfy c  dðeÞ < u  c  v < c þ dðeÞ, then we have, j f ðvÞ  f ðuÞ  ðv  uÞ f 0 ðcÞj  eðv  uÞ. [Hint: The dðeÞ is given by Definition 6.1.1. Subtract, and add the term f ðcÞ  cf 0 ðcÞ on the left side and use the Triangle Inequality.]

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C06, , 12/09/2010, , 14:54:0, , 172, , Page 172, , CHAPTER 6 DIFFERENTIATION, , Section 6.2 The Mean Value Theorem, The Mean Value Theorem, which relates the values of a function to values of its derivative,, is one of the most useful results in real analysis. In this section we will establish this, important theorem and sample some of its many consequences., We begin by looking at the relationship between the relative extrema of a function and, the values of its derivative. Recall that the function f : I ! R is said to have a relative, maximum [respectively, relative minimum] at c 2 I if there exists a neighborhood, V :¼ V d ðcÞ of c such that f ðxÞ  f ðcÞ [respectively, f ðcÞ  f ðxÞ] for all x in V \ I. We, say that f has a relative extremum at c 2 I if it has either a relative maximum or a relative, minimum at c., The next result provides the theoretical justification for the familiar process of finding, points at which f has relative extrema by examining the zeros of the derivative. However, it, must be realized that this procedure applies only to interior points of the interval. For, example, if f ðxÞ :¼ x on the interval I :¼ ½0; 1, then the endpoint x ¼ 0 yields the unique, relative minimum and the endpoint x ¼ 1 yields the unique maximum of f on I, but neither, point is a zero of the derivative of f., 6.2.1 Interior Extremum Theorem Let c be an interior point of the interval I at which, f : I ! R has a relative extremum. If the derivative of f at c exists, then f 0 ðcÞ ¼ 0., Proof. We will prove the result only for the case that f has a relative maximum at c; the, proof for the case of a relative minimum is similar., If f 0 ðcÞ > 0, then by Theorem 4.2.9 there exists a neighborhood V
I of c such that, f ðxÞ  f ðcÞ, > 0 for, xc, If x 2 V and x > c, then we have, , x 2 V; x 6¼ c:, , f ðxÞ  f ðcÞ, > 0:, xc, But this contradicts the hypothesis that f has a relative maximum at c. Thus we cannot, have f 0 ðcÞ > 0. Similarly (how?), we cannot have f 0 ðcÞ < 0. Therefore we must have, f 0 ðcÞ ¼ 0., Q.E.D., f ð xÞ  f ð c Þ ¼ ð x  c Þ , , 6.2.2 Corollary Let f : I ! R be continuous on an interval I and suppose that f has a, relative extremum at an interior point c of I. Then either the derivative of f at c does not, exist, or it is equal to zero., We note that if f ðxÞ :¼ jxj on I :¼ ½1; 1, then f has an interior minimum at x ¼ 0;, however, the derivative of f fails to exist at x ¼ 0., 6.2.3 Rolle’s Theorem Suppose that f is continuous on a closed interval I :¼ ½a; b, that, the derivative f 0 exists at every point of the open interval (a, b), and that f ðaÞ ¼ f ðbÞ ¼ 0., Then there exists at least one point c in (a, b) such that f 0 ðcÞ ¼ 0., Proof. If f vanishes identically on I, then any c in (a, b) will satisfy the conclusion of the, theorem. Hence we suppose that f does not vanish identically; replacing f by f if, necessary, we may suppose that f assumes some positive values. By the MaximumMinimum Theorem 5.3.4, the function f attains the value supf f ðxÞ : x 2 I g > 0 at some, point c in I. Since f ðaÞ ¼ f ðbÞ ¼ 0, the point c must lie in (a, b); therefore f 0 ðcÞ exists.

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C06, , 12/09/2010, , 14:54:1, , Page 173, , 6.2 THE MEAN VALUE THEOREM, , Figure 6.2.1, , 173, , Rolle’s Theorem, , Since f has a relative maximum at c, we conclude from the Interior Extremum, Theorem 6.2.1 that f 0 ðcÞ ¼ 0. (See Figure 6.2.1.), Q.E.D., As a consequence of Rolle’s Theorem, we obtain the fundamental Mean Value, Theorem., 6.2.4 Mean Value Theorem Suppose that f is continuous on a closed interval, I :¼ ½a; b, and that f has a derivative in the open interval (a, b). Then there exists at, least one point c in (a, b) such that, f ðbÞ  f ðaÞ ¼ f 0 ðcÞðb  aÞ:, Proof. Consider the function w defined on I by, wðxÞ :¼ f ðxÞ  f ðaÞ , , f ð bÞ  f ð aÞ, ðx  aÞ:, ba, , [The function w is simply the difference of f and the function whose graph is the line, segment joining the points (a, f (a)) and (b, f (b)); see Figure 6.2.2.] The hypotheses of, , Figure 6.2.2, , The Mean Value Theorem

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C06, , 12/09/2010, , 14:54:2, , 174, , Page 174, , CHAPTER 6 DIFFERENTIATION, , Rolle’s Theorem are satisfied by w since w is continuous on [a, b], differentiable on (a, b),, and wðaÞ ¼ wðbÞ ¼ 0. Therefore, there exists a point c in (a, b) such that, 0 ¼ w 0 ð cÞ ¼ f 0 ð cÞ , , f ðbÞ  f ðaÞ, :, ba, , Hence, f ðbÞ  f ðaÞ ¼ f 0 ðcÞðb  aÞ., , Q.E.D., , Remark The geometric view of the Mean Value Theorem is that there is some point on, the curve y ¼ f ðxÞ at which the tangent line is parallel to the line segment through the, points (a, f (a)) and (b, f (b)). Thus it is easy to remember the statement of the Mean Value, Theorem by drawing appropriate diagrams. While this should not be discouraged, it tends, to suggest that its importance is geometrical in nature, which is quite misleading. In fact the, Mean Value Theorem is a wolf in sheep’s clothing and is the Fundamental Theorem of, Differential Calculus. In the remainder of this section, we will present some of the, consequences of this result. Other applications will be given later., The Mean Value Theorem permits one to draw conclusions about the nature of a, function f from information about its derivative f 0 . The following results are obtained in, this manner., 6.2.5 Theorem Suppose that f is continuous on the closed interval I :¼ ½a; b, that f is, differentiable on the open interval (a, b), and that f 0 ðxÞ ¼ 0 for x 2 ða; bÞ. Then f is, constant on I., Proof. We will show that f ðxÞ ¼ f ðaÞ for all x 2 I. Indeed, if x 2 I, x > a, is given,, we apply the Mean Value Theorem to f on the closed interval ½a, x. We obtain a, point c (depending on x) between a and x such that f ðxÞ  f ðaÞ ¼ f 0 ðcÞðx  aÞ. Since, f 0 ðcÞ ¼ 0 (by hypothesis), we deduce that f ðxÞ  f ðaÞ ¼ 0. Hence, f ðxÞ ¼ f ðaÞ for, any x 2 I., Q.E.D., 6.2.6 Corollary Suppose that f and g are continuous on I :¼ ½a; b, that they are, differentiable on (a, b), and that f 0 ðxÞ ¼ g0 ðxÞ for all x 2 ða; bÞ. Then there exists a, constant C such that f ¼ g þ C on I., Recall that a function f : I ! R is said to be increasing on the interval I if whenever, x1 ; x2 in I satisfy x1 < x2, then f ðx1 Þ  f ðx2 Þ. Also recall that f is decreasing on I if the, function  f is increasing on I., 6.2.7 Theorem Let f : I ! R be differentiable on the interval I. Then:, (a) f is increasing on I if and only if f 0 ðxÞ  0 for all x 2 I., (b) f is decreasing on I if and only if f 0 ðxÞ  0 for all x 2 I., Proof. (a) Suppose that f 0 ðxÞ  0 for all x 2 I. If x1 ; x2 in I satisfy x1 < x2, then we, apply the Mean Value Theorem to f on the closed interval J :¼ ½x1 ; x2  to obtain a point c in, ðx1 ; x2 Þ such that, f ðx2 Þ  f ðx1 Þ ¼ f 0 ðcÞðx2  x1 Þ:

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C06, , 12/09/2010, , 14:54:3, , Page 175, , 6.2 THE MEAN VALUE THEOREM, , 175, , Since f 0 ðcÞ  0 and x2  x1 > 0, it follows that f ðx2 Þ  f ðx1 Þ  0. (Why?) Hence,, f ðx1 Þ  f ðx2 Þ and, since x1 < x2 are arbitrary points in I, we conclude that f is, increasing on I., For the converse assertion, we suppose that f is differentiable and increasing on I., Thus, for any point x 6¼ c in I, we have ð f ðxÞ  f ðcÞÞ=ðx  cÞ  0. (Why?) Hence, by, Theorem 4.2.6 we conclude that, f 0 ðcÞ ¼ lim, , x!c, , f ðxÞ  f ðcÞ,  0:, xc, , (b) The proof of part (b) is similar and will be omitted., , Q.E.D., , A function f is said to be strictly increasing on an interval I if for any points x1 ; x2 in I, such that x1 < x2 , we have f ðx1 Þ < f ðx2 Þ. An argument along the same lines of the proof, of Theorem 6.2.7 can be made to show that a function having a strictly positive derivative, on an interval is strictly increasing there. (See Exercise 13.) However, the converse, assertion is not true, since a strictly increasing differentiable function may have a derivative, that vanishes at certain points. For example, the function f : R ! R defined by f ðxÞ :¼ x3, is strictly increasing on R, but f 0 ð0Þ ¼ 0. The situation for strictly decreasing functions is, similar., Remark It is reasonable to define a function to be increasing at a point if there is a, neighborhood of the point on which the function is increasing. One might suppose that, if, the derivative is strictly positive at a point, then the function is increasing at this point., However, this supposition is false; indeed, the differentiable function defined by, gðxÞ :¼, , x þ 2x2 sin ð1=xÞ, 0, , if, if, , x 6¼ 0;, x ¼ 0;, , is such that g0 ð0Þ ¼ 1, yet it can be shown that g is not increasing in any neighborhood of, x ¼ 0. (See Exercise 10.), We next obtain a sufficient condition for a function to have a relative extremum at an, interior point of an interval., 6.2.8 First Derivative Test for Extrema Let f be continuous on the interval I :¼ ½a; b, and let c be an interior point of I. Assume that f is differentiable on (a, c) and (c, b) . Then:, (a), and, (b), and, , If there is a neighborhood ðc  d; c þ dÞ
I such that f 0 ðxÞ  0 for c  d < x < c, f 0 ðxÞ  0 for c < x < c þ d, then f has a relative maximum at c., If there is a neighborhood ðc  d; c þ dÞ
I such that f 0 ðxÞ  0 for c  d < x < c, f 0 ðxÞ  0 for c < x < c þ d, then f has a relative minimum at c., , Proof. (a) If x 2 ðc  d; cÞ, then it follows from the Mean Value Theorem that there, exists a point cx 2 ðx; cÞ such that f ðcÞ  f ðxÞ ¼ ðc  xÞ f 0 ðcx Þ. Since f 0 ðcx Þ  0 we infer, that f ðxÞ  f ðcÞ for x 2 ðc  d; cÞ. Similarly, it follows (how?) that f ðxÞ  f ðcÞ, for x 2 ðc; c þ dÞ. Therefore f ðxÞ  f ðcÞ for all x 2 ðc  d; c þ dÞ so that f has a relative, maximum at c., (b) The proof is similar., Q.E.D., Remark The converse of the First Derivative Test 6.2.8 is not true. For example, there, exists a differentiable function f : R ! R with absolute minimum at x ¼ 0 but such that

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C06, , 12/09/2010, , 14:54:3, , 176, , Page 176, , CHAPTER 6 DIFFERENTIATION, , f 0 takes on both positive and negative values on both sides of (and arbitrarily close to), x ¼ 0. (See Exercise 9.), Further Applications of the Mean Value Theorem, We will continue giving other types of applications of the Mean Value Theorem; in doing, so we will draw more freely than before on the past experience of the reader and his or her, knowledge concerning the derivatives of certain well-known functions., 6.2.9 Examples (a) Rolle’s Theorem can be used for the location of roots of a function., For, if a function g can be identified as the derivative of a function f, then between any two, roots of f there is at least one root of g. For example, let gðxÞ :¼ cos x, then g is known to be, the derivative of f ðxÞ :¼ sin x. Hence, between any two roots of sin x there is at least one, root of cos x. On the other hand, g0 ðxÞ ¼ sin x ¼ f ðxÞ, so another application of Rolle’s, Theorem tells us that between any two roots of cos there is at least one root of sin., Therefore, we conclude that the roots of sin and cos interlace each other. This conclusion is, probably not news to the reader; however, the same type of argument can be applied to the, Bessel functions Jn of order n ¼ 0; 1; 2; . . . by using the relations, 0, , ½xn J n ðxÞ0 ¼ xn J n1 ðxÞ; ½xn1 J n ðxÞ ¼ xn J nþ1 ðxÞ, , for, , x > 0:, , The details of this argument should be supplied by the reader., (b) We can apply the Mean Value Theorem for approximatepcalculations, and to obtain, ffiffiffiffiffiffiffiffi, error estimates. For example, suppose, it, is, desired, to, evaluate, 105, ., We, employ, the Mean, pffiffiffi, Value Theorem with f ðxÞ :¼ x; a ¼ 100; b ¼ 105, to obtain, pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi, 5, 105  100 ¼ pffiffiffi, 2 c, pffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi, for some number c with 100 < c < 105. Since 10 < c < 105 < 121 ¼ 11, we can, assert that, pffiffiffiffiffiffiffiffi, 5, 5, < 105  10 <, ;, 2ð11Þ, 2ð10Þ, pffiffiffiffiffiffiffiffi, whence it follows that 10:2272 < 105p, <ffiffiffi 10:2500., estimate, may not be as sharp as, pffiffiffiffiffiffiffiffi This, pffiffiffiffiffiffiffi, ffi, desired. It is clear that the estimate c < 105, <, 121, was, wasteful, can be, pffiffiffiffiffiffiffiffi, pffiffiand, ffi, improved by making use of our conclusion that 105 < 10:2500. Thus, c < 10:2500, and we easily determine that, pffiffiffiffiffiffiffiffi, 5, < 105  10:, 2ð10:2500Þ, pffiffiffiffiffiffiffiffi, Our improved estimate is 10:2439 < 105 < 10:2500., 0:2439 <, , &, , Inequalities, One very important use of the Mean Value Theorem is to obtain certain inequalities., Whenever information concerning the range of the derivative of a function is available, this, information can be used to deduce certain properties of the function itself. The following, examples illustrate the valuable role that the Mean Value Theorem plays in this respect.

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C06, , 12/09/2010, , 14:54:3, , Page 177, , 6.2 THE MEAN VALUE THEOREM, , 177, , 6.2.10 Examples (a) The exponential function f ðxÞ :¼ ex has the derivative f 0 ðxÞ ¼ ex, for all x 2 R. Thus f 0 ðxÞ > 1 for x > 0, and f 0 ðxÞ < 1 for x < 0. From these relationships,, we will derive the inequality, ð1Þ, , ex  1 þ x for, , x 2 R;, , with equality occurring if and only if x ¼ 0., If x ¼ 0, we have equality with both sides equal to 1. If x > 0, we apply the Mean, Value Theorem to the function f on the interval [0, x]. Then for some c with 0 < c < x, we have, ex  e0 ¼ ec ðx  0Þ:, Since e0 ¼ 1 and ec > 1, this becomes ex  1 > x so that we have ex > 1 þ x for x > 0. A, similar argument establishes the same strict inequality for x < 0. Thus the inequality (1), holds for all x, and equality occurs only if x ¼ 0., (b) The function gðxÞ :¼ sin x has the derivative g0 ðxÞ ¼ cos x for all x 2 R. On the basis, of the fact that 1  cos x  1 for all x 2 R, we will show that, ð2Þ, , x  sin x  x, , for all x  0:, , Indeed, if we apply the Mean Value Theorem to g on the interval [0, x], where x > 0, we, obtain, sin x  sin 0 ¼ ðcos cÞðx  0Þ, for some c between 0 and x. Since sin 0 ¼ 0 and 1  cos c  1, we have x  sin x  x., Since equality holds at x ¼ 0, the inequality (2) is established., (c) (Bernoulli’s inequality) If a > 1, then, ð3Þ, , ð1 þ xÞa  1 þ ax, , for all, , x > 1;, , with equality if and only if x ¼ 0., This inequality was established earlier, in Example 2.1.13(c), for positive integer, values of a by using Mathematical Induction. We now derive the more general version by, employing the Mean Value Theorem., If hðxÞ :¼ ð1 þ xÞa then h0 ðxÞ ¼ að1 þ xÞa1 for all x > 1. [For rational a this, derivative was established in Example 6.1.10(c). The extension to irrational will be, discussed in Section 8.3.] If x > 0, we infer from the Mean Value Theorem applied to, h on the interval [0, x] that there exists c with 0 < c < x such that hðxÞ  hð0Þ ¼ h0 ðcÞðx  0Þ., Thus, we have, ð1 þ xÞa  1 ¼ að1 þ cÞa1 x:, Since c > 0 and a  1 > 0, it follows that ð1 þ cÞa1 > 1 and hence that ð1 þ xÞa >, 1 þ ax. If 1 < x < 0, a similar use of the Mean Value Theorem on the interval [x, 0] leads, to the same strict inequality. Since the case x ¼ 0 results in equality, we conclude that (3) is, valid for all x > 1 with equality if and only if x ¼ 0., (d) Let a be a real number satisfying 0 < a < 1 and let gðxÞ ¼ ax  xa for x  0. Then, g0 ðxÞ ¼ að1  xa1 Þ, so that g0 ðxÞ < 0 for 0 < x < 1 and g0 ðxÞ > 0 for x > 1. Consequently, if x  0, then gðxÞ  gð1Þ and gðxÞ ¼ gð1Þ if and only if x ¼ 1. Therefore, if, x  0 and 0 < a < 1, then we have, xa  a x þ ð1  aÞ:

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C06, , 12/09/2010, , 14:54:4, , 178, , Page 178, , CHAPTER 6 DIFFERENTIATION, , If a > 0 and b > 0 and if we let x ¼ a=b and multiply by b, we obtain the inequality, aa b1a  aa þ ð1  aÞb;, &, , where equality holds if and only if a ¼ b., The Intermediate Value Property of Derivatives, , We conclude this section with an interesting result, often referred to as Darboux’s Theorem. It, states that if a function f is differentiable at every point of an interval I, then the function f 0 has, the Intermediate Value Property. This means that if f 0 takes on values A and B, then it also, takes on all values between A and B. The reader will recognize this property as one of the, important consequences of continuity as established in Theorem 5.3.7. It is remarkable that, derivatives, which need not be continuous functions, also possess this property., 6.2.11 Lemma Let I
R be an interval, let f : I ! R, let c 2 I, and assume that f has a, derivative at c. Then:, (a) If f 0 ðcÞ > 0, then there is a number d > 0 such that f ðxÞ > f ðcÞ for x 2 I such that, c < x < c þ d., (b) If f 0 ðcÞ < 0, then there is a number d > 0 such that f ðxÞ > f ðcÞ for x 2 I such that, c  d < x < c., Proof. (a) Since, f ðxÞ  f ðcÞ, ¼ f 0 ðcÞ > 0;, xc, it follows from Theorem 4.2.9 that there is a number d > 0 such that if x 2 I and, 0 < jx  cj < d, then, lim, , x!c, , f ðxÞ  f ðcÞ, > 0:, xc, If x 2 I also satisfies x > c, then we have, f ðxÞ  f ðcÞ, > 0:, xc, Hence, if x 2 I and c < x < c þ d, then f ðxÞ > f ðcÞ., The proof of (b) is similar., f ðxÞ  f ðcÞ ¼ ðx  cÞ , , Q.E.D., , 6.2.12 Darboux’s Theorem If f is differentiable on I ¼ ½a; b and if k is a number, between f 0 ðaÞ and f 0 ðbÞ, then there is at least one point c in (a, b) such that f 0 ðcÞ ¼ k., Proof. Suppose that f 0 ðaÞ < k < f 0 ðbÞ. We define g on I by gðxÞ :¼ kx  f ðxÞ for x 2 I., Since g is continuous, it attains a maximum value on I. Since g0 ðaÞ ¼ k  f 0 ðaÞ > 0, it, follows from Lemma 6.2.11(a) that the maximum of g does not occur at x ¼ a. Similarly,, since g0 ðbÞ ¼ k  f 0 ðbÞ < 0, it follows from Lemma 6.2.11(b) that the maximum does not, occur at x ¼ b. Therefore, g attains its maximum at some c in (a, b). Then from Theorem, 6.2.1 we have 0 ¼ g0 ðcÞ ¼ k  f 0 ðcÞ. Hence, f 0 ðcÞ ¼ k., Q.E.D., 6.2.13 Example The function g: ½1; 1 ! R, 8, for, < 1, gðxÞ :¼, 0, for, :, 1, for, , defined by, 0 < x  1;, x ¼ 0;, 1  x < 0;

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C06, , 12/09/2010, , 14:54:4, , Page 179, , 6.2 THE MEAN VALUE THEOREM, , 179, , (which is a restriction of the signum function) clearly fails to satisfy the intermediate value, property on the interval ½1; 1. Therefore, by Darboux’s Theorem, there does not exist a, function f such that f 0 ðxÞ ¼ gðxÞ for all x 2 ½1; 1. In other words, g is not the derivative, &, on ½1; 1 of any function., Exercises for Section 6.2, 1. For each of the following functions on R to R, find points of relative extrema, the intervals on, which the function is increasing, and those on which it is decreasing:, (a) f ðxÞ :¼ x2  3x þ 5;, (b) gðxÞ :¼ 3x  4x2 ;, 3, (c) hðxÞ :¼ x  3x  4;, (d) kðxÞ :¼ x4 þ 2x2  4:, 2. Find the points of relative extrema, the intervals on which the following functions are, increasing, and those on which they are decreasing:, (a) f ðxÞ :¼ x þ 1=x for x 6¼ 0;, (b) gðxÞ :¼ x=ðx2 þ 1Þ for x 2 R;, pffiffiffiffiffiffiffiffiffiffiffi, pffiffiffi, (c) hðxÞ :¼ x  2 x þ 2 for x > 0;, (d) kðxÞ :¼ 2x þ 1=x2 for x 6¼ 0:, 3. Find the points, extrema of the following functions on the specified domain:,
of relative,
, (b) gðxÞ :¼ 1  ðx  1Þ2=3 for 0  x  2;, (a) f ðxÞ :¼
x2  1
for  4  x  4;,
2,
, (d) kðxÞ :¼ xðx  8Þ1=3 for 0  x  9:, (c) hðxÞ :¼ x
x  12
for  2  x  3;, 4. Let a1 ; a2 ; . . . ; an be real numbers and let f be defined on R by, n, X, ðai  xÞ2 for x 2 R:, f ðxÞ :¼, i¼1, , Find the unique point of relative minimum for f., 5. Let a > b > 0 and let n 2 N satisfy n  2. Prove that a1=n  b1=n < ða  bÞ1=n . [Hint: Show that, f ðxÞ :¼ x1=n  ðx  1Þ1=n is decreasing for x  1, and evaluate f at 1 and a=b.], 6. Use the Mean Value Theorem to prove that jsin x  sin yj  jx  yj for all x, y in R., 7. Use the Mean Value Theorem to prove that ðx  1Þ=x < ln x < x  1 for x > 1. [Hint: Use the, fact that D ln x ¼ 1=x for x > 0.], 8. Let f : ½a; b ! R be continuous on [a, b] and differentiable in (a, b). Show that if lim f 0 ðxÞ ¼ A,, x!a, , then f 0 ðaÞ exists and equals A. [Hint: Use the definition of f 0 ðaÞ and the Mean Value Theorem.], 9. Let f : R ! R be defined by f ðxÞ :¼ 2x4 þ x4 sinð1=xÞ for x 6¼ 0 and f ð0Þ :¼ 0. Show that f, has an absolute minimum at x ¼ 0, but that its derivative has both positive and negative values in, every neighborhood of 0., 10. Let g : R ! R be defined by gðxÞ :¼ x þ 2x2 sinð1=xÞ for x 6¼ 0 and gð0Þ :¼ 0. Show that, g0 ð0Þ ¼ 1, but in every neighborhood of 0 the derivative g0 ðxÞ takes on both positive and, negative values. Thus g is not monotonic in any neighborhood of 0., 11. Give an example of a uniformly continuous function on [0,1] that is differentiable on (0, 1) but, whose derivative is not bounded on (0, 1)., 12. If hðxÞ :¼ 0 for x < 0 and hðxÞ :¼ 1 for x  0, prove there does not exist a function f : R ! R, such that f 0 ðxÞ ¼ hðxÞ for all x 2 R. Give examples of two functions, not differing by a, constant, whose derivatives equal h(x) for all x 6¼ 0., 13. Let I be an interval and let f : I ! R be differentiable on I. Show that if f 0 is positive on I, then f, is strictly increasing on I., 14. Let I be an interval and let f : I ! R be differentiable on I. Show that if the derivative f 0 is never, 0 on I, then either f 0 ðxÞ > 0 for all x 2 I or f 0 ðxÞ < 0 for all x 2 I., 15. Let I be an interval. Prove that if f is differentiable on I and if the derivative f 0 is bounded on I,, then f satisfies a Lipschitz condition on I. (See Definition 5.4.4.)

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C06, , 12/09/2010, , 14:54:5, , 180, , Page 180, , CHAPTER 6 DIFFERENTIATION, , 16. Let f : ½0; 1Þ ! R be differentiable on (0, 1) and assume that f 0 ðxÞ ! b as x ! 1., (a) Show that for any h > 0, we have lim ð f ðx þ hÞ  f ðxÞÞ=h ¼ b., x!1, , (b) Show that if f ðxÞ ! a as x ! 1, then b ¼ 0., (c) Show that lim ð f ðxÞ=xÞ ¼ b., x!1, , 17. Let f, g be differentiable on R and suppose that f ð0Þ ¼ gð0Þ and f 0 ðxÞ  g0 ðxÞ for all x  0., Show that f ðxÞ  gðxÞ for all x  0., 18. Let I :¼ ½a; b and let f : I ! R be differentiable at c 2 I. Show that for every e > 0 there exists, d > 0 such that if 0 < jx  yj < d and a  x  c  y  b, then,
,
,
f ðxÞ  f ðyÞ,
, 0,
,
,
x  y  f ðcÞ
< e:, 19. A differentiable function f : I ! R is said to be uniformly differentiable on I :¼ ½a; b if for, every e > 0 there exists d > 0 such that if 0 < jx  yj < d and x; y 2 I, then,
,
,
,
f ðx Þ  f ðy Þ, 0,
< e:,
, , f, ð, x, Þ,
,
xy, Show that if f is uniformly differentiable on I, then f 0 is continuous on I., 20. Suppose that f : ½0; 2 ! R is continuous on [0, 2] and differentiable on (0, 2), and that, f ð0Þ ¼ 0; f ð1Þ ¼ 1; f ð2Þ ¼ 1., (a) Show that there exists c1 2 ð0; 1Þ such that f 0 ðc1 Þ ¼ 1., (b) Show that there exists c2 2 ð1; 2Þ such that f 0 ðc2 Þ ¼ 0., (c) Show that there exists c 2 ð0; 2Þ such that f 0 ðcÞ ¼ 1=3., , Section 6.3 L’Hospital’s Rules, In this section we will discuss limit theorems that involve cases that cannot be determined, by previous limit theorems. For example, if f (x) and g(x) both approach 0 as x approaches, a, then the quotient f ðxÞ=gðxÞ may or may not have a limit at a and it is said to have the, indeterminate form 0/0. The limit theorem for this case is due to Johann Bernoulli and first, appeared in the 1696 book published by L’Hospital., , Johann Bernoulli, Johann Bernoulli (1667–1748) was born in Basel, Switzerland. Johann, worked for a year in his father’s spice business, but he was not a success., He enrolled in Basel University to study medicine, but his brother Jacob,, twelve years older and a Professor of Mathematics, led him into mathematics., Together, they studied the papers of Leibniz on the new subject of calculus., Johann received his doctorate at Basel University and joined the faculty at, Groningen in Holland, but upon Jacob’s death in 1705, he returned to Basel, and was awarded Jacob’s chair in mathematics. Because of his many advances, in the subject, Johann is regarded as one of the founders of calculus., While in Paris in 1692, Johann met the Marquis Guillame Francois de L’Hospital and agreed, to a financial arrangement under which he would teach the new calculus to L’Hospital, giving, L’Hospital the right to use Bernoulli’s lessons as he pleased. This was subsequently continued, through a series of letters. In 1696, the first book on differential calculus, L’Analyse des Infiniment, Petits, was published by L’Hospital. Though L’Hospital’s name was not on the title page, his, portrait was on the frontispiece and the preface states ‘‘I am indebted to the clarifications of the, brothers Bernoulli, especially the younger.’’ The book contains a theorem on limits later known as, L’Hospital’s Rule although it was in fact discovered by Johann Bernoulli. In 1922, manuscripts, were discovered that confirmed the book consisted mainly of Bernoulli’s lessons. And in 1955,, the L’Hospital–Bernoulli correspondence was published in Germany.

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C06, , 12/09/2010, , 14:54:5, , Page 181, , 6.3 L’HOSPITAL’S RULES, , 181, , The initial theorem was refined and extended, and the various results are collectively, referred to as L’Hospital’s (or L’H^, opital’s) Rules. In this section we establish the most basic, of these results and indicate how others can be derived., Indeterminate Forms, In the preceding chapters we have often been concerned with methods of evaluating limits. It, was shown in Theorem 4.2.4(b) that if A :¼ lim f ðxÞ and B :¼ lim gðxÞ, and if B 6¼ 0, then, x!c, , lim, , f ðxÞ, , x!c gðxÞ, , ¼, , x!c, , A, :, B, , However, if B ¼ 0, then no conclusion was deduced. It will be seen in Exercise 2 that if, B ¼ 0 and A 6¼ 0, then the limit is infinite (when it exists)., The case A ¼ 0, B ¼ 0 has not been covered previously. In this case, the limit of the, quotient f=g is said to be ‘‘indeterminate.’’ We will see that in this case the limit may not, exist or may be any real value, depending on the particular functions f and g. The, symbolism 0=0 is used to refer to this situation. For example, if a is any real number, and if, we define f ðxÞ :¼ ax and gðxÞ :¼ x, then, f ð xÞ, ax, lim, ¼ lim, ¼ lim a ¼ a:, x!0 gðxÞ, x!0 x, x!0, Thus the indeterminate form 0=0 can lead to any real number a as a limit., Other indeterminate forms are represented by the symbols 1=1; 0  1;, 00 ; 11 ; 10 ; and 1  1. These notations correspond to the indicated limiting behavior, and juxtaposition of the functions f and g. Our attention will be focused on the indeterminate, forms 0=0 and 1=1. The other indeterminate cases are usually reduced to the form 0=0 or, 1=1 by taking logarithms, exponentials, or algebraic manipulations., A Preliminary Result, To show that the use of differentiation in this context is a natural and not surprising, development, we first establish an elementary result that is based simply on the definition, of the derivative., 6.3.1 Theorem Let f and g be defined on [a, b], let f ðaÞ ¼ gðaÞ ¼ 0, and let gðxÞ 6¼ 0 for, a < x < b. If f and g are differentiable at a and if g0 ðaÞ 6¼ 0, then the limit of f=g at a, exists and is equal to f 0 ðaÞ=g0 ðaÞ. Thus, f ðxÞ f 0 ðaÞ, ¼ 0 :, lim, x!aþ gðxÞ, g ðaÞ, Proof. Since f ðaÞ ¼ gðaÞ ¼ 0, we can write the quotient f ðxÞ=gðxÞ for a < x < b, as follows:, f ðxÞ  f ðaÞ, f ðxÞ f ðxÞ  f ðaÞ, xa, ¼, ¼, gðxÞ gðxÞ  gðaÞ gðxÞ  gðaÞ, :, xa, Applying Theorem 4.2.4(b), we obtain, f ðxÞ  f ðaÞ, lim, f ðxÞ x!aþ, f 0 ð aÞ, xa, lim, ¼, ¼ 0 :, Q.E.D., x!aþ gðxÞ, gðxÞ  gðaÞ g ðaÞ, lim, x!aþ, xa

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C06, , 12/09/2010, , 14:54:5, , 182, , Page 182, , CHAPTER 6 DIFFERENTIATION, , Warning The hypothesis that f ðaÞ ¼ gðaÞ ¼ 0 is essential here. For example, if f ðxÞ, :¼ x þ 17 and gðxÞ :¼ 2x þ 3 for x 2 R, then, lim, , x!0, , f ðxÞ 17, ¼ ;, gðxÞ, 3, , while, , f 0 ð0Þ 1, ¼ :, g0 ð 0Þ 2, , The preceding result enables us to deal with limits such as, x2 þ x 2  0 þ 1 1, ¼, ¼ :, x!0 sin 2x, 2 cos 0, 2, , lim, , To handle limits where f and g are not differentiable at the point a, we need a more general, version of the Mean Value Theorem due to Cauchy., 6.3.2 Cauchy Mean Value Theorem Let f and g be continuous on [a, b] and, differentiable on (a, b), and assume that g0 ðxÞ 6¼ 0 for all x in (a, b). Then there exists, c in (a, b) such that, f ðbÞ  f ðaÞ f 0 ðcÞ, ¼, :, gðbÞ  gðaÞ g0 ðcÞ, Proof. As in the proof of the Mean Value Theorem, we introduce a function to which, Rolle’s Theorem will apply. First we note that since g0 ðxÞ 6¼ 0 for all x in (a, b), it follows, from Rolle’s Theorem that gðaÞ 6¼ gðbÞ. For x in [a, b], we now define, hðxÞ :¼, , f ð bÞ  f ð aÞ, ðgðxÞ  gðaÞÞ  ð f ðxÞ  f ðaÞÞ:, gð bÞ  gð aÞ, , Then h is continuous on [a, b], differentiable on (a, b), and hðaÞ ¼ hðbÞ ¼ 0. Therefore, it, follows from Rolle’s Theorem 6.2.3 that there exists a point c in (a, b) such that, 0 ¼ h0 ð c Þ ¼, , f ð bÞ  f ð aÞ 0, g ðcÞ  f 0 ðcÞ:, gð bÞ  gð aÞ, , Since g0 ðcÞ 6¼ 0, we obtain the desired result by dividing by g0 ðcÞ., , Q.E.D., , Remarks The preceding theorem has a geometric interpretation that is similar to that of, the Mean Value Theorem 6.2.4. The functions f and g can be viewed as determining a curve, in the plane by means of the parametric equations x ¼ f ðtÞ; y ¼ gðtÞ where a  t  b., Then the conclusion of the theorem is that there exists a point ð f ðcÞ; gðcÞÞ on the curve for, some c in (a, b) such that the slope g0 ðcÞ=f 0 ðcÞ of the line tangent to the curve at that point is, equal to the slope of the line segment joining the endpoints of the curve., Note that if gðxÞ ¼ x, then the Cauchy Mean Value Theorem reduces to the Mean, Value Theorem 6.2.4., , L’Hospital’s Rule, I, We will now establish the first of L’Hospital’s Rules. For convenience, we will consider, right-hand limits at a point a; left-hand limits, and two-sided limits are treated in exactly, the same way. In fact, the theorem even allows the possibility that a ¼ 1. The reader, should observe that, in contrast with Theorem 6.3.1, the following result does not assume, the differentiability of the functions at the point a. The result asserts that the limiting, behavior of f ðxÞ=gðxÞ as x ! aþ is the same as the limiting behavior of f 0 ðxÞ=g0 ðxÞ as

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C06, , 12/09/2010, , 14:54:6, , 184, , Page 184, , CHAPTER 6 DIFFERENTIATION, , are equal. In the examples that follow, we will apply the appropriate version of L’Hospital’s, Rule as needed., 6.3.4 Examples (a) We have, , , , pffiffiffi, sin x, cos x, pffiffiffi ¼ lim 2 x cos x ¼ 0:, lim pffiffiffi ¼ lim, x!0þ, x!0þ 1=ð2 xÞ, x!0þ, x, , Observe that the denominator is not differentiable, pffiffiffiat x ¼ 0 so that Theorem 6.3.1, cannot be applied. However f ðxÞ :¼ sin x and gðxÞ :¼ x are differentiable on (0, 1) and, both approach 0 as x ! 0þ. Moreover, g0 ðxÞ 6¼ 0 on (0, 1), so that 6.3.3 is applicable., , , 1  cos x, sin x, :, (b) We have lim, ¼ lim, x!0, x!0 2x, x2, The quotient in the second limit is again indeterminate in the form 0=0. However, the, hypotheses are again satisfied so that a second application of L’Hospital’s Rule is, permissible. Hence, we obtain, , , 1  cos x, sin x, cos x 1, lim, ¼ lim, ¼ :, ¼ lim, x!0, x!0 2x, x!0 2, x2, 2, ex  1, ex, ¼ lim ¼ 1., x!0, x!0 1, x, Similarly, two applications of L’Hospital’s Rule give us,  x, , e 1x, ex  1, ex 1, ¼ lim ¼ :, lim, ¼ lim, 2, x!0, x!0 2x, x!0 2, x, 2, , , ln x, ð1=xÞ, ¼ 1., (d) We have lim, ¼ lim, x!1 x  1, x!1, 1, (c) We have lim, , &, , L’Hospital’s Rule, II, This rule is very similar to the first one, except that it treats the case where the denominator, becomes infinite as x ! aþ. Again we will consider only right-hand limits, but it is, possible that a ¼ 1. Left-hand limits and two-sided limits are handled similarly., 6.3.5 L’Hospital’s Rule, II Let 1  a < b  1 and let f, g be differentiable on (a, b), such that g0 ðxÞ 6¼ 0 for all x 2 ða; bÞ . Suppose that, ð5Þ, , lim gðxÞ ¼, , x!aþ, , 1:, , f 0 ðxÞ, f ðxÞ, ¼ L., ¼ L 2 R, then lim, x!aþ g0 ðxÞ, x!aþ gðxÞ, , (a) If lim, , f 0 ðxÞ, f ðxÞ, ¼ L., ¼ L 2 f1; 1g, then lim, x!aþ g0 ðxÞ, x!aþ gðxÞ, , (b) If lim, , Proof. We will suppose that (5) holds with limit 1., As before, we have gðbÞ 6¼ gðaÞ for a; b 2 ða; bÞ; a < b. Further, equation (2) in the, proof of 6.3.3 holds for some u 2 ða; bÞ., Case (a): If L 2 R with L > 0 and e > 0 is given, there is c 2 ða; bÞ such that (3) in the, proof of 6.3.3 holds when a < a < b  c. Since gðxÞ ! 1, we may also assume that

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C06, , 12/09/2010, , 14:54:7, , 186, , Page 186, , CHAPTER 6 DIFFERENTIATION, , Here we take f ðxÞ :¼ x2 and gðxÞ :¼ ex on R. We obtain, x2, 2x, 2, ¼ lim x ¼ lim x ¼ 0:, x!1 ex, x!1 e, x!1 e, ln sin x, ., (c) We consider lim, x!0þ ln x, Here we take f ðxÞ :¼ ln sin x and gðxÞ :¼ ln x on ð0; pÞ. If we apply 6.3.5, we obtain, h x i, ln sin x, cos x=sin x, lim, ¼ lim, ¼ lim,  ½cos x:, x!0þ ln x, x!0þ, x!0þ sin x, 1=x, lim, , Since lim ½x=sin x ¼ 1 and lim cos x ¼ 1, we conclude that the limit under considerax!0þ, , x!0þ, , tion equals 1., x  sin x, . This has indeterminate form 1=1. An application of, x!1 x þ sin x, L’Hospital’s Rule gives us, 1  cos x, ;, lim, x!1 1 þ cos x, which is useless because this limit does not exist. (Why not?) However, if we rewrite the, original limit, we get directly that, (d) Consider lim, , sin x, x ¼ 1  0 ¼ 1:, lim, x!1, sin x 1 þ 0, 1þ, x, 1, , &, , Other Indeterminate Forms, Indeterminate forms such as 1  1; 0  1; 11 ; 00 ; 10 can be reduced to the previously, considered cases by algebraic manipulations and the use of the logarithmic and exponential, functions. Instead of formulating these variations as theorems, we illustrate the pertinent, techniques by means of examples., 6.3.7 Examples (a) Let I :¼ ð0; p=2Þ and consider, , , 1, 1, , lim, ;, x!0þ x, sin x, which has the indeterminate form 1  1. We have, , , 1, 1, sin x  x, cos x  1, , ¼ lim, ¼ lim, lim, x!0þ x, x!0þ, x!0þ, sin x, x sin x, sin x þ x cos x, sin x, 0, ¼ ¼ 0:, ¼ lim, x!0þ 2 cos x  x sin x, 2, (b) Let I :¼ ð0; 1Þ and consider lim x ln x, which has the indeterminate form, x!0þ, 0  ð1Þ. We have, lim x ln x ¼ lim, , x!0þ, , ln x, , x!0þ 1=x, , ¼ lim, , 1=x, , x!0þ 1=x2, , ¼ lim ðxÞ ¼ 0:, x!0þ, , (c) Let I :¼ ð0; 1Þ and consider lim xx, which has the indeterminate form 00., x!0þ, , We recall from calculus (see also Section 8.3) that xx ¼ ex ln x . It follows from part (b), and the continuity of the function y 7! ey at y ¼ 0 that lim xx ¼ e0 ¼ 1., x!0þ

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C06, , 12/09/2010, , 14:54:9, , 188, , Page 188, , CHAPTER 6 DIFFERENTIATION, , 9. Evaluate the following limits:, ln x, (a) lim 2, ð0; 1Þ,, x!1 x, (c), , (b), , lim x ln sin x ð0; pÞ,, , (d), , x!0, , ln x, lim pffiffiffi ð0; 1Þ,, x, x þ ln x, ð0; 1Þ., lim, x!1 x ln x, x!1, , 10. Evaluate the following limits:, (a), (c), , lim x2x, , x!0þ, , ð0; 1Þ,, , lim ð1 þ 3=xÞx, , x!1, , (b), , ð0; 1Þ,, , (d), , lim ð1 þ 3=xÞx ð0; 1Þ,, , , 1, 1, , ð0; 1Þ., lim, x!0þ x, Arctan x, x!0, , 11. Evaluate the following limits:, (a), (c), , lim x1=x, , x!1, , lim xsin x, , x!0þ, , ð0; 1Þ,, , (b), , ð0; 1Þ,, , (d), , lim ðsin xÞx, , x!0þ, , ð0; pÞ,, , lim ðsec x  tan xÞ ð0; p=2Þ., , x!p=2, , 12. Let f be differentiable on (0, 1) and suppose that lim ð f ðxÞ þ f 0 ðxÞÞ ¼ L. Show that, x!1, , lim f ðxÞ ¼ L and lim f 0 ðxÞ ¼ 0. [Hint: f ðxÞ ¼ ex f ðxÞ=ex .], x!1, x!1, tan x, as x ! ðp=2Þ. Then evaluate directly, 13. Try to use L’Hospital’s Rule to find the limit of, sec x, by changing to sines and cosines., 14. Show that if c > 0, then lim, , x!c, , xc  cx, 1  ln c, ., ¼, 1 þ ln c, xx  cc, , Section 6.4 Taylor’s Theorem, A very useful technique in the analysis of real functions is the approximation of functions, by polynomials. In this section we will prove a fundamental theorem in this area that goes, back to Brook Taylor (1685–1731), although the remainder term was not provided until, much later by Joseph-Louis Lagrange (1736–1813). Taylor’s Theorem is a powerful result, that has many applications. We will illustrate the versatility of Taylor’s Theorem by briefly, discussing some of its applications to numerical estimation, inequalities, extreme values of, a function, and convex functions., Taylor’s Theorem can be regarded as an extension of the Mean Value Theorem to, ‘‘higher order’’ derivatives. Whereas the Mean Value Theorem relates the values of a, function and its first derivative, Taylor’s Theorem provides a relation between the values of, a function and its higher order derivatives., Derivatives of order greater than one are obtained by a natural extension of the, differentiation process. If the derivative f 0 ðxÞ of a function f exists at every point x in an, interval I containing a point c, then we can consider the existence of the derivative of the, function f 0 at the point c. In case f 0 has a derivative at the point c, we refer to the resulting, number as the second derivative of f at c, and we denote this number by f 00 ðcÞ or by, f (2)(c). In similar fashion we define the third derivative f 000 ðcÞ ¼ f ð3Þ ðcÞ; . . ., and the nth, derivative f (n)(c), whenever these derivatives exist. It is noted that the existence of the, nth derivative at c presumes the existence of the (n  l)st derivative in an interval, containing c, but we do allow the possibility that c might be an endpoint of such an, interval., If a function f has an nth derivative at a point x0, it is not difficult to construct an, ðkÞ, nth degree polynomial Pn such that Pn ðx0 Þ ¼ f ðx0 Þ and PðkÞ, ðx0 Þ for, n ðx0 Þ ¼ f

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C06, , 12/09/2010, , 14:54:10, , 190, , Page 190, , CHAPTER 6 DIFFERENTIATION, , Rn ðxÞ where Rn is given by, Rn ðxÞ :¼, , ð3Þ, , f ðnþ1Þ ðcÞ, ðx  x0 Þnþ1, ðn þ 1Þ!, , for some point c between x and x0. This formula for Rn is referred to as the Lagrange form, (or the derivative form) of the remainder. Many other expressions for Rn are known; one is, in terms of integration and will be given later. (See Theorem 7.3.18.), , Applications of Taylor’s Theorem, The remainder term Rn in Taylor’s Theorem can be used to estimate the error in, approximating a function by its Taylor polynomial Pn. If the number n is prescribed,, then the question of the accuracy of the approximation arises. On the other hand, if a certain, accuracy is specified, then the question of finding a suitable value of n is germane. The, following examples illustrate how one responds to these questions., pffiffiffiffiffiffiffiffiffiffiffiffiffi, 6.4.2 Examples (a) Use Taylor’s Theorem with n ¼ 2 to approximate 3 1 þ x; x > 1., We take the function f ðxÞ :¼ ð1 þ xÞ1=3 , the point x0 ¼ 0, and n ¼ 2. Since f 0 ðxÞ ¼, 2=3, 1, and f 00 ðxÞ ¼ 13 ð 23Þð1 þ xÞ5=3 , we have f 0 ð0Þ ¼ 13 and f 00 ð0Þ ¼ 2=9. Thus, 3 ð1 þ xÞ, we obtain, f ðxÞ ¼ P2 ðxÞ þ R2 ðxÞ ¼ 1 þ 13 x  19 x2 þ R2 ðxÞ;, 5, ð1 þ cÞ8=3 x3 for some point c between 0 and x., where R2 ðxÞ ¼ 3!1 f 000 ðcÞx3 ¼ 81, pffiffiffiffiffiffiffiffi, For example, if we let x ¼ 0.3, we get the approximation P2(0.3) ¼ 1.09 for 3 1:3:, Moreover, since c > 0 in this case, then ð1 þ cÞ8=3 < 1 and so the error is at most,  , 5 3 3, 1, < 0:17 102 :, R2 ð0:3Þ , ¼, 81 10, 600, pffiffiffiffiffiffiffi, Hence, we have j 3 1:31:09j < 0:5 102 , so that two decimal place accuracy is assured., (b) Approximate the number e with error less than 105 ., We shall consider the function gðxÞ :¼ ex and take x0 ¼ 0 and x ¼ 1 in Taylor’s, Theorem. We need to determine n so that jRn ð1Þj < 105 . To do so, we shall use the fact, that g0 ðxÞ ¼ ex and the initial bound of ex  3 for 0  x  1., Since g0 ðxÞ ¼ ex , it follows that gðkÞ ðxÞ ¼ ex for all k 2 N, and therefore gðkÞ ð0Þ ¼ 1, for all k 2 N. Consequently the nth Taylor polynomial is given by, , x2, xn, þ  þ, 2!, n!, and the remainder for x ¼ 1 is given by Rn ð1Þ ¼ ec =ðn þ 1Þ! for some c satisfying 0 < c < 1., Since ec < 3, we seek a value of n such that 3=ðn þ 1Þ! < 105 . A calculation reveals that, 9! ¼ 362; 880 > 3 105 so that the value n ¼ 8 will provide the desired accuracy; moreover,, since 8! ¼ 40; 320, no smaller value of n will be certain to suffice. Thus, we obtain, Pn ðxÞ :¼ 1 þ x þ, , e, , P8 ð1Þ ¼ 1 þ 1 þ, , 1, 1, þ    þ ¼ 2:718 28, 2!, 8!, , with error less than 105 ., Taylor’s Theorem can also be used to derive inequalities., , &

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C06, , 12/09/2010, , 14:54:10, , Page 191, , 6.4 TAYLOR’S THEOREM, , 191, , 6.4.3 Examples (a) 1  12 x2  cos x for all x 2 R., Use f ðxÞ :¼ cos x and x0 ¼ 0 in Taylor’s Theorem, to obtain, 1, cos x ¼ 1  x2 þ R2 ðxÞ;, 2, where for some c between 0 and x we have, f 000 ðcÞ 3 sin c 3, x ¼, x :, R2 ðxÞ ¼, 3!, 6, If 0  x  p, then 0  c < p; since c and x3 are both positive, we have R2 ðxÞ  0. Also,, if p  x  0, then p  c  0; since sin c and x3 are both negative, we again have, R2 ðxÞ  0. Therefore, we see that 1  12 x2  cos x for jxj  p. If jxj  p, then we have, 1  12 x2 < 3  cos x and the inequality is trivially valid. Hence, the inequality holds for, all x 2 R., (b) For any k 2 N, and for all x > 0, we have, 1, 1, 1, 1, x2kþ1 :, x  x2 þ     x2k < lnð1 þ xÞ < x  x2 þ    þ, 2, 2k, 2, 2k þ 1, Using the fact that the derivative of ln(l þ x) is 1=(1 þ x) for x > 0, we see that the nth, Taylor polynomial for ln(l þ x) with x0 ¼ 0 is, 1, 1, Pn ðxÞ ¼ x  x2 þ    þ ð1Þn1 xn, 2, n, and the remainder is given by, ð1Þn cnþ1 nþ1, x, nþ1, for some c satisfying 0 < c < x. Thus for any x > 0, if n ¼ 2k is even, then we have, R2k ðxÞ > 0; and if n ¼ 2k þ 1 is odd, then we have R2kþ1 ðxÞ < 0. The stated inequality, then follows immediately., (c) ep > pe, Taylor’s Theorem gives us the inequality ex > 1 þ x for x > 0, which the reader, should verify. Then, since p > e, we have x ¼ p=e  1 > 0, so that, Rn ðxÞ ¼, , This implies ep=e, , eðp=e1Þ > 1 þ ðp=e  1Þ ¼ p=e:, > ðp=eÞe ¼ p, and thus we obtain the inequality ep > pe., , &, , Relative Extrema, It was established in Theorem 6.2.1 that if a function f : I ! R is differentiable at a point c, interior to the interval I, then a necessary condition for f to have a relative extremum at c is, that f 0 ðcÞ ¼ 0. One way to determine whether f has a relative maximum or relative, minimum [or neither] at c is to use the First Derivative Test 6.2.8. Higher order derivatives,, if they exist, can also be used in this determination, as we now show., 6.4.4 Theorem Let I be an interval, let x0 be an interior point of I, and let n  2., Suppose that the derivatives f 0 ; f 00 . . . ; f ðnÞ exist and are continuous in a neighborhood, of x0 and that f 0 ðx0 Þ ¼    ¼ f ðn1Þ ðx0 Þ ¼ 0, but f ðnÞ ðx0 Þ 6¼ 0., (i), (ii), (iii), , If n is even and f ðnÞ ðx0 Þ > 0, then f has a relative minimum at x0., If n is even and f ðnÞ ðx0 Þ < 0, then f has a relative maximum at x0., If n is odd, then f has neither a relative minimum nor relative maximum at x0.

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C06, , 12/09/2010, , 14:54:11, , 192, , Page 192, , CHAPTER 6 DIFFERENTIATION, , Proof. Applying Taylor’s Theorem at x0, we find that for x 2 I we have, f ðnÞ ðcÞ, ðx  x0 Þn ;, n!, where c is some point between x0 and x. Since f (n) is continuous, if f ðnÞ ðx0 Þ 6¼ 0, then there, exists an interval U containing x0 such that f ðnÞ ðxÞ will have the same sign as f ðnÞ ðx0 Þ for, x 2 U. If x 2 U, then the point c also belongs to U and consequently f ðnÞ ðcÞ and f ðnÞ ðx0 Þ, will have the same sign., f ðxÞ ¼ Pn1 ðxÞ þ Rn1 ðxÞ ¼ f ðx0 Þ þ, , (i) If n is even and f ðnÞ ðx0 Þ > 0, then for x 2 U we have f ðnÞ ðcÞ > 0 and ðx  x0 Þn , 0 so that Rn1 ðxÞ  0. Hence, f ðxÞ  f ðx0 Þ for x 2 U, and therefore f has a relative, minimum at x0., (ii) If n is even and f ðnÞ ðx0 Þ < 0, then it follows that Rn1 ðxÞ  0 for x 2 U, so that, f ðxÞ  f ðx0 Þ for x 2 U. Therefore, f has a relative maximum at x0., (iii) If n is odd, then ðx  x0 Þn is positive if x > x0 and negative if x < x0 . Consequently, if x 2 U, then Rn1 ðxÞ will have opposite signs to the left and to the right of x0., Therefore, f has neither a relative minimum nor a relative maximum at x0 ., Q.E.D., Convex Functions, The notion of convexity plays an important role in a number of areas, particularly in the, modern theory of optimization. We shall briefly look at convex functions of one real, variable and their relation to differentiation. The basic results, when appropriately, modified, can be extended to higher dimensional spaces., 6.4.5 Definition Let I
R be an interval. A function f : I ! R is said to be convex on I, if for any t satisfying 0  t  1 and any points x1 ; x2 in I, we have, f ðð1  tÞx1 þ tx2 Þ  ð1  tÞ f ðx1 Þ þ tf ðx2 Þ:, Note that if x1 < x2 , then as t ranges from 0 to 1, the point ð1  tÞx1 þ tx2 traverses, the interval from x1 to x2 . Thus if f is convex on I and if x1 ; x2 2 I, then the chord joining, any two points ðx1 ; f ðx1 ÞÞ and ðx2 ; f ðx2 ÞÞ on the graph of f lies above the graph of f. (See, Figure 6.4.1.), , Figure 6.4.1, , A convex function, , A convex function need not be differentiable at every point, as the example, f ðxÞ :¼ jxj; x 2 R, reveals. However, it can be shown that if I is an open interval and if, f : I ! R is convex on I, then the left and right derivatives of f exist at every point of I. As a

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C06, , 12/09/2010, , 14:54:12, , 194, , Page 194, , CHAPTER 6 DIFFERENTIATION, , Let f be a differentiable function that has a zero at r and let x1 be an initial estimate of, r. The line tangent to the graph at ðx1 ; f ðx1 ÞÞ has the equation y ¼ f ðx1 Þ þ f 0 ðx1 Þðx  x1 Þ,, and crosses the x-axis at the point, x2 :¼ x1 , , f ðx1 Þ, :, f 0 ðx1 Þ, , (See Figure 6.4.2.) If we replace x1 by the second estimate x2 , then we obtain a point x3 ,, and so on. At the nth iteration we get the point xnþ1 from the point xn by the formula, xnþ1 :¼ xn , , f ðxn Þ, :, f 0 ðxn Þ, , Under suitable hypotheses, the sequence ðxn Þ will converge rapidly to a root of the equation, f ðxÞ ¼ 0, as we now show. The key tool in establishing the rapid rate of convergence is, Taylor’s Theorem., , Figure 6.4.2, , Newton’s Method, , 6.4.7 Newton’s Method Let I :¼ ½a; b and let f : I ! R be twice differentiable on I., Suppose that f ðaÞ f ðbÞ < 0 and that there are constants m, M such that j f 0 ðxÞj  m > 0, and j f 00 ðxÞj  M for x 2 I and let K :¼ M=2m. Then there exists a subinterval I, containing a zero r of f such that for any x1 2 I the sequence ðxn Þ defined by, ð5Þ, , xnþ1 :¼ xn , , f ðxn Þ, f 0 ðxn Þ, , f or all, , n 2 N;, , belongs to I and ðxn Þ converges to r. Moreover, ð6Þ, , jxnþ1  rj  K jxn  rj2, , f or all, , n 2 N:, , Proof. Since f ðaÞ f ðbÞ < 0, the numbers f (a) and f (b) have opposite signs; hence by, Theorem 5.3.5 there exists r 2 I such that f ðrÞ ¼ 0. Since f 0 is never zero on I, it follows, from Rolle’s Theorem 6.2.3 that f does not vanish at any other point of I., We now let x0 2 I be arbitrary; by Taylor’s Theorem there exists a point c0 between x0, and r such that, 0 ¼ f ðrÞ ¼ f ðx0 Þ þ f 0 ðx0 Þðr  x0 Þ þ 12 f 00 ðc0 Þðr  x0 Þ2 ;

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C06, , 12/09/2010, , 14:54:14, , 196, , Page 196, , CHAPTER 6 DIFFERENTIATION, , are illustrated in Figures 6.4.3 and 6.4.4. One familiar strategy is to use the Bisection, Method to arrive at a fairly close estimate of the root and then to switch to Newton’s, Method for the coup de gr^ace., , Figure 6.4.3, , Figure 6.4.4 xn oscillates, between x1 and x2, , xn ! 1, , Exercises for Section 6.4, 1. Let f ðxÞ :¼ cos ax for x 2 R where a 6¼ 0. Find f ðnÞ ðxÞ for n 2 N, x 2 R., 2. Let gðxÞ :¼ jx3 j for x 2 R. Find g0 ðxÞ and g00 ðxÞ for x 2 R, and g000 ðxÞ for x 6¼ 0. Show that, g000 ð0Þ does not exist., 3. Use Induction to prove Leibniz’s rule for the nth derivative of a product:, n  , X, n ðnkÞ, f, ðxÞgðkÞ ðxÞ:, ð f gÞðnÞ ðxÞ ¼, k, k¼0, , pffiffiffiffiffiffiffiffiffiffiffi, 1 þ x  1 þ 12 x., pffiffiffiffiffiffiffi, pffiffiffi, 5. Use the preceding exercise to approximate 1:2 and 2. What is the best accuracy you can be, sure of, using this inequality?, pffiffiffiffiffiffiffi, pffiffiffi, 6. Use Taylor’s Theorem with n ¼ 2 to obtain more accurate approximations for 1:2 and 2., 4. Show that if x > 0, then 1 þ 12 x  18 x2 , , 1=3, 1, 1 2, 3, 7. If x >p0ffiffiffiffiffiffi, show, that, p, ffiffiffijð1 þ xÞ  ð1 þ 3 x  9 x Þj  ð5=81Þx . Use this inequality to approxiffi, 3, 3, mate 1:2 and 2., , 8. If f ðxÞ :¼ ex , show that the remainder term in Taylor’s Theorem converges to zero as n ! 1,, for each fixed x0 and x. [Hint: See Theorem 3.2.11.], 9. If gðxÞ :¼ sin x, show that the remainder term in Taylor’s Theorem converges to zero as n ! 1, for each fixed x0 and x., 10. Let hðxÞ :¼ e1=x for x 6¼ 0 and hð0Þ :¼ 0. Show that hðnÞ ð0Þ ¼ 0 for all n 2 N. Conclude that, the remainder term in Taylor’s Theorem for x0 ¼ 0 does not converge to zero as n ! 1 for, x 6¼ 0. [Hint: By L’Hospital’s Rule, lim hðxÞ=xk ¼ 0 for any k 2 N. Use Exercise 3 to calculate, x!0, hðnÞ ðxÞ for x 6¼ 0.], 2, , 11. If x 2 ½0; 1 and n 2 N, show that,
, , 
, 2, 3, n
, nþ1,
,
lnð1 þ xÞ  x  x þ x þ    þ ð1Þn1 x
< x, :,
, 2, 3, n
nþ1, Use this to approximate ln 1.5 with an error less than 0.01. Less than 0.001.

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C06, , 12/09/2010, , 14:54:14, , Page 197, , 6.4 TAYLOR’S THEOREM, , 197, , 12. We wish to approximate sine by a polynomial on ½1; 1 so that the error is less than 0.001., Show that we have,
, , 
, 3, 5,
,
,
sin x  x  x þ x
< 1, for jxj  1:,
, 6 120
5040, 13. Calculate e correct to seven decimal places., 14. Determine whether or not x ¼ 0 is a point of relative extremum of the following functions:, (a) f ðxÞ :¼ x3 þ 2;, (b) gðxÞ :¼ sin x  x;, (c), , hðxÞ :¼ sin x þ 16 x3 ;, , (d), , kðxÞ :¼ cos x  1 þ 12 x2 :, , 15. Let f be continuous on [a, b] and assume the second derivative f 00 exists on (a, b). Suppose that, the graph of f and the line segment joining the points ða; f ðaÞÞ and ðb; f ðbÞÞ intersect at a point, ðx0 ; f ðx0 ÞÞ where a < x0 < b. Show that there exists a point c 2 ða; bÞ such that f 00 ðcÞ ¼ 0., 16. Let I
R be an open interval, let f : I ! R be differentiable on I, and suppose f 00 ðaÞ exists at, a 2 I. Show that, f 00 ðaÞ ¼ lim, , h!0, , f ða þ hÞ  2 f ðaÞ þ f ða  hÞ, :, h2, , Give an example where this limit exists, but the function does not have a second derivative at a., 17. Suppose that I
R is an open interval and that f 00 ðxÞ  0 for all x 2 I. If c 2 I, show that the, part of the graph of f on I is never below the tangent line to the graph at (c, f (c))., 18. Let I
R be an interval and let c 2 I. Suppose that f and g are defined on I and that the, derivatives f ðnÞ ; gðnÞ exist and are continuous on I. If f ðkÞ ðcÞ ¼ 0 and gðkÞ ðcÞ ¼ 0 for, k ¼ 0; 1; . . . ; n  1, but gðnÞ ðcÞ 6¼ 0, show that, lim, , x!c, , f ðxÞ f ðnÞ ðcÞ, ¼, :, gðxÞ gðnÞ ðcÞ, , 19. Show that the function f ðxÞ :¼ x3  2x  5 has a zero r in the interval I :¼ ½2; 2:2. If x1 :¼ 2, and if we define the sequence (xn) using the Newton procedure, show that, jxnþ1  rj  ð0:7Þjxn  rj2 . Show that x4 is accurate to within six decimal places., 20. Approximate the real zeros of gðxÞ :¼ x4  x  3., 21. Approximate the real zeros of hðxÞ :¼ x3  x  1. Apply Newton’s Method starting with the, initial choices (a) x1 :¼ 2, (b) x1 :¼ 0, (c) x1 :¼ 2. Explain what happens., 22. The equation ln x ¼ x  2 has two solutions. Approximate them using Newton’s Method. What, happens if x1 :¼ 12 is the initial point?, 23. The function f ðxÞ ¼ 8x3  8x2 þ 1 has two zeros in [0,1]. Approximate them, using Newton’s, Method, with the starting points (a) x1 :¼ 18, (b) x1 :¼ 14. Explain what happens., 24. Approximate the solution of the equation x ¼ cos x, accurate to within six decimals.