(Solution Manual) Instructor’s Solution Manual Introduction to Real Analysis

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Instructor’s Solution Manual INTRODUCTION TO REAL ANALYSIS (Comprehensive Guide ) Contents Chapter 1 The Real Numbers 1 1.1 The Real Number System 1 1.2 Mathematical Induction 4 1.3 The Real Line 13 Chapter 2 Differential Calculus of Functions of One Variable 17 2.1 Functions and Limits 17 2.2 Continuity 24 2.3 Differentiable Functions of One Variable 30 2.4 L’Hospital’s Rule 36 2.5 Taylor’s Theorem 43 Chapter 3 Integral Calculus of Functions of One Variable 53 3.1 Definition of the Integral 53 3.2 Existence of the Integral 56 3.3 Properties of the Integral 61 3.4 Improper Integrals 66 3.5 A More Advanced Look at the Existence of the Proper Riemann Integral 77 Chapter 4 Infinite Sequences and Series 79 4.1 Sequences of Real Numbers 79 4.2 Earlier Topics Revisited With Sequences 87 4.3 Infinite Series of Constants 89 4.4 Sequences and Series of Functions 100 4.5 Power Series 107 Chapter 5 Real -Valued Functions of Several Variables 116 5.1 Structure of Rn 116 5.2 Continuous Real -Valued Function of n Variables 121 5.3 Partial Derivatives and the Differential 123 5.4 The Chain Rule and Taylor’s Theorem 130 Chapter 6 Vector -Valued Functions of Several Variables 141 6.1 Linear Transformations and Matrices 141 6.2 Continuity and Differentiability of Transformations 146 6.3 The Inverse Function Theorem 152 6.4 The Implicit Function Theorem 160 Chapter 7 Integrals of Functions of Several Variables 170 7.1 Definition and Existence of the Multiple Integral 170 7.2 Iterated Integrals and Multiple Integrals 187 7.3 Change of Variables in Multiple Integrals 207 Chapter 8 Metric Spaces 217 8.1 Introduction to Metric Spaces 217 8.2 Compact Sets in a Metric Space 224 8.3 Continuous Functions on Metric Spaces 226 Section 1.1 The Real Number System 1 ˇ ˇ ˇ ˇ D D D D D D D C C D C D C C D C D D C C D D D C C D C C C D C C C D C C Cj — jC C — Cj — j D C C j — j D D C C —j — j— C — —j — j D C — j — j D D D C D C D CHAPTER 1 THE REAL NUMBERS 1.1 THE REAL NUMBER SYSTEM 1:1:1 . Note that ja — bj D max .a; b/ — min.a; b/. (a) a C b C ja — bj D a C b C max .a; b/ — min.a; b/ D 2 max .a; b/. (b) a C b — ja — bj D a C b — max .a; b/ C min.a; b/ D 2 min.a; b/. (c) Let ˛ a b 2c a b a b 2c a b . From (a), ˛ 2 Œmax.a; b/ c max.a; b/ c ] df ˇ. From (a) with a and b replaced by max .a; b/ and c, ˇ 4 max .max .a; b/; c/ 4 max .a; b; c/. (d) Let ˛ a b 2c a b a b 2c a b . From (b), ˛ 2 Œmin.a; b/ c min.a; b/ c ] df ˇ. From (a) with a and b replaced by min.a; b/ and c, ˇ 4 min .min.a; b/; c/ 4 min.a; b; c/. 1:1:2 . First verify axioms A-E: Axiom A. See Eqns. (1.1.1) and (1.1.2). Axiom B. If a 0 then .a b/ c b c and a .b c/ b c, so .a b/ c a .b c/. Similar arguments apply if b 0 or c 0. The remaining case is a b c 1. Since .1 1/ 1 0 1 1 and 1 .1 1/ 1 0 1, addition is associative. Since .ab/c a.bc/ 0; unless a D b D c D 1; 1; if a D b D c D 1; multiplication is associative. Axiom C. Since a.b c/ ab ac 0; if a D 0; b C c; if a D 1; the distributive law holds. Axiom D. Eqns. (1.1.1) and (1.1.2) imply that 0 and 1 have the required properties. 2 Chapter 1 The Real Numbers 1 C C C C 1 ≥ 2 ≥ C D C 2 ≤ ≤ ≤ 2 2 2 2 — — D C C — 1 1 Axiom E. Eqn. (1.1.1) implies that —0 D 0 and —1 D 1; Eqn. (1.1.2) implies that 1=1 D 1. To see that the field cannot be ordered suppose that 0 < 1. Adding 1 to both sides yields 0 1 < 1 1, or 1 < 0, a contradiction. On the other hand, if 1 < 0, then 1 1 < 0 1, so 0 < 1, also a contradiction. 1:1:3 . If p 2 is rational we can write p 2 D m=n, where either m is odd or n is odd. Then m2 D 2n2, so m is even; thus, m D 2m 1 where m1 is an integer. Therefore, 4m2 D 2n2, so n2 D 2m2 and n is also even, a contradiction. 1:1:4 . If pp is rational we can write pp D m=n where either m or n is not divisible by p. Then m2 D pn2, so m is divisible by p; thus, m D pm 1 where m1 is an integer. Therefore, p2m2 D pn2, so n2 D pm2 and n is also divisible by p, a contradiction. 1:1:6 . If S is bounded below then T is bounded above, so T has a unique supremum, by Theorem 1.1.3. Denote sup T D —˛. Then (i) if x 2 S then —x ≤ —˛, so x ≥ ˛; (ii) if ‹ > 0 there is an x0 2 T such that —x0 < —˛ — ‹, so x0 > ˛ C ‹. Therefore, there is an ˛ with properties (i) and (ii). If (i) and (ii) hold with ˛ replaced by ˛1 then —˛1 is a supremum of T , so ˛1 D ˛ by the uniqueness assertion of Theorem 1.1.3. 1:1:7 . (a) If x S, then inf S x sup S, and the transitivity of implies (A), with equality if and only if S contains exactly one point. (b) There are three cases: (i) If S is bounded below and unbounded above, then inf S D ˛ (finite) and sup S D 1 from (13); (ii) If S is unbounded below and bounded above, then inf S D —1 from (14) and sup S D ˇ (finite); (iii) If S is unbound ed below and above then sup S D 1 from (13) and inf S D —1 from (14). In all three cases (12) implies (A). 1:1:8 . Let a D inf T and b D sup S. We first show that a D b. If a < b then a < .a C b/=2 < b. Since b D sup S, there is an s0 2 S such that s0 > .a C b/=2 . Since a D inf T , there is a t0 2 T such that t0 < .a C b/=2 . Therefore, t0 < s0, a contradiction. Hence a ≥ b. If a > b there is an x such that b < x < a . From the definitions of a and b, x 62 T and x 62 S, a contradiction. Hence a D b. Let ˇ D a.D b/. Since a and b are uniquely, defined so is ˇ. If x > ˇ then x … S (since ˇ D sup S), so x 2 T . If x < ˇ then x … T (since ˇ D inf T ), so x 2 S. 1:1:9 . Every real number is in either S or T . T is nonempty because U is bounded. S is nonempty because if u U and x < u then x S. If s S there is a u U such that u > s, since s is not an upper bound of U . If t T then t u , since t is an upper bound of U . Since u > s and t u, t > s. Therefore, S and T satisfy the conditions imposed in Exercise 1.1.8, so there is a number ˇ such that every number greater than ˇ is an upper bound of U and no number less that ˇ is an upper bound of U . However, ˇ is also an upper bound of U (if not, there would be a u0 2 U such that u0 > ˇ, which is impossible, since if u0 > ˇ then every number in .ˇ; u0/ is an upper bound of U ). Therefore, ˇ D sup U . 1:1:10 . (a) Let ˛ D sup S and ˇ D sup T . If x D s C t then x ≤ ˛ C ˇ. If ‹ > 0 choose s0 in S and t0 in T such that s0 > ˛ ‹=2 and t0 > ˇ ‹=2. Then x0 s0 t0 > ˛ ˇ ‹ and ˛ ˇ sup.S T / by Theorem 1.1.3. This proves (A). The proof of (B) by mean of Theorem 1.1.8 is similar. (b) For (A), suppose that S is not bounded above. Then sup S D sup.S C T / D 1. Since sup T > —1 if T is nonempty, (A) holds. The proof of (B) is similar. Section 1.1 The Real Number System 3 1:1:11 . Apply Exercise 1.1.10 with S and T replaced by S and —T . 1:1:12 . If a D 0 then T D fbg, so inf T D sup T D b. Now suppose that a ¤ 0 and let ˛ D inf S and ˇ D sup S. From the definitions of ˛ and ˛ ≤ s and s ≤ ˇ for all s 2 S; .A/ and if ‹ > 0 is given there are elements s1 and s2 of S such that s1 < ˛ C ‹=jaj and s2 > ˇ — ‹=jaj: .B/ CASE 2. If a > 0, multiplying the inequalities in (A) by a shows that a˛ ≤ as and as ≤ aˇ for all s 2 S: Therefore, a˛ C b ≤ as C b for all s 2 S; .C/ as C b ≤ aˇ C b for all s 2 S: .D/ Multiplying the inequalities in (B) by a shows that as1 < a˛ C ‹ and as2 > aˇ — ‹; since jaj D a. Therefore, as1 C b < .a˛ C b/ C ‹; .E/ as2 C b > .aˇ C b/ — ‹: .F/ Now (C) and (E) imply that a˛ C b D inf T , while (D) and (F) imply that aˇ C b D sup T . CASE 3. Suppose that a < 0. Multiplying the inequalities in (A) by a shows that a˛ ≥ as and as ≥ aˇ for all s 2 S: Therefore, a˛ C b ≥ as C b for all s 2 S; .G/ as C b ≥ aˇ C b for all s 2 S: .H/ Multiplying the inequalities in (B) by a shows that as1 > a˛ — ‹ and as2 < aˇ C ‹; since jaj D —a. Therefore, as1 C b > .a˛ C b/ — ‹; .I/ as2 C b < .aˇ C b/ C ‹: .J/ Now (G) and (I) imply that a˛ C b D sup T , while (H) and (J) imply that aˇ C b D inf T . ˇ, 4 Chapter 1 The Real Numbers 2 2 3 3 ≥ 1.2 MATHEMATICAL INDUCTION 1:2:1 . P1 is obvious. If Pn is true, then 1 C 3 C · · · C .2n C 1/ D Œ1 C 3 C · · · .2n — 1/] C .2n C 1/ D n C .2n C 1/ (by Pn) D .n C 1/ ; so PnC1 is true. 1:2:2 . P1 is obvious. If Pn is true, then 12 C 22 C · · · C .n C 1/2 D Œ12 C 22 C · · · C n2] C .n C 1/2 D n.n C 1/.2n C 1/ C .n C 1/2 (by P ) 6 n .n C 1/.n C 2/.2n C 3/ ; 6 so PnC1 is true. 1:2:3 . Since 1 D 1, P1 is true. Suppose that n ≥ 1 and Pn is true. Then 12 C 32 C · · · C .2n C 1/2 D 12 C 32 C · · · C .2n — 1/2 C .2n C 1/2 n.4n2 — 1/ 2 D 3 C .2n C 1/ (by Pn) D .2n C 1/ n.2n — 1/ C 2n C 1 D 2n C 1 Œ2n2 C 5n C 3] .n C 1/.2n C 1/.2n C 3/ 3 .n C 1/Œ2.n C 1/ — 1]Œ2.n C 1/ C 1] 3 .n C 1/Œ4.n C 1/2 — 1] that is, D 2 2 2 3 I nŒ4.n C 1/2 — 1/ 1 C 3 C · · · C .2n C 1/ D 3 ; which is PnC1. 1:2:4 . P1 is trivial. P2 is true by the triangle inequality (Theorem 1.1.1). If n 2 and Pn is true then ja1 C a2 C · · · C an C anC1j D j.a1 C a2 C · · · C an/ C anC1j ≤ ja1 C a2 C · · · C anj C janC1j .A/ D D D Section 1.2 Mathematical Induction 5 2 h ı i C by the triangle inequality with a D a1 C a2 C · · · C an and b D anC1. By Pn, ja1 C a2 C · · · C anj ≤ ja1j C ja2j C · · · C janj: This and (A) imply PnC1. 1:2:5 . P1 is obvious. If Pn is true then .1 C a1/.1 C a2/ · · · .1 C anC1/ D Œ.1 C a1/ · · · .1 C an/].1 C anC1/ ≥ .1 C a1 C · · · C an/.1 C anC1/ (by Pn) D 1 C a1 C · · · C anC1 C anC1.a1 C · · · C an/; which implies PnC1, since ai ≥ 0. 1:2:6 . P1 is obvious. If Pn is true then .1 — a1/.1 — a2/ · · · .1 — anC1/ ≥ .1 — anC1/.1 — a1 — · · · — an/ (by Pn) D 1 — a1 — · · · — anC1 C anC1.a1 C · · · C an/; which implies PnC1. 1:2:7 . If sn > 0 then 0 < e—sn < 1, so 0 < snC1 < 1. Therefore P1 is true and Pn implies PnC1. 1:2:8 . Subtracting p R D 1 R p R C p R from the equation defining xn 1 yields (A) xnC1 — p R D .xn — p R/2=.2x n/. Since x0 > 0, this implies by induction that xn > p R for all n ≥ 1, and the definitipon of xnC1 now impplies that xnC1 < xn if n ≥ 1. Now let Pn n 2 be the proposition that xnp— R ≤ 2— .x0 — R/ =x0. Settingpn D 0 in (A) vperifies P1. n D 0. Since 0 < .xn — Pn implies PnC1. R/=x n < 1, (A) implies that xnC1 — R < .xn — R/=2 , so 1:2:9 . (a) Rewrite the formula as anC1 2an D .2n C 1/.2n C 2/: We compute a few terms to formulate Pn: 2a1 22 a2 D 3 · 4 D 4Š ; 2a2 2 22 23 a3 D 5 · 6 D 5 · 6 4Š D 6Š ; 2a3 2 23 24 Let Pn be the proposition that a4 D 7 · 8 D 7 · 8 6Š D 8Š ; 2n an D .2n/Š ; 6 Chapter 1 The Real Numbers · · · 2 2.2/2 1 2.2/2 which is true for n D 1 (also for n D 2; 3; 4). If Pn is true then 2an 2 2n 2nC1 anC1 D .2n C 1/.2n C 2/ D .2n C 1/.2n C 2/ .2n/Š D .2n C 2/Š; so Pn implies PnC1. (b) We compute a few terms to formulate Pn: 3a1 3 a2 D 5 D 4 · 5 ; 3a2 3 3 32 a3 D 7 · 6 D 6 · 7 4 · 5 D 4 · 5 · 6 · 7 ; 3a3 3 32 33 a4 D 8 · 9 D 8 · 9 4 · 5 · 6 · 7 D 4 · 5 · 6 · 7 · 8 · 9 : Multiply numerator and denominator by 3Š D 6 in each fraction yields simpler results: a2 D Let Pn to be proposition that 2 32 5Š ; a3 D 2 32 7Š a4 D 2 · 3n 2 34 : 9Š an D .2n C 1/Š ; which is true for n D 1 (also for n D 2; 3; 4). If Pn is true then 3an 3 2 · 3n 2 · 3nC1 anC1 D .2n C 3/.2n C 2/ D .2n C 3/.2n C 2/ .2n C 1/Š D .2n C 3/Š; so Pn implies PnC1. (c) Rewrite the formula as anC1 .2n C 1/.2n C 2/a n : 2.n C 1/2 We compute a few terms to formulate Pn: a D 3 · 4 a D 3 · 4 ; a D 5 · 6 a D 5 · 6 3 · 4 D 3 · 4 · 5 · 6 ; 3 2.3/2 2 2.3/2 2.2/2 22.2 · 3/2 a D 7 · 8 a D 7 · 8 3 · 4 · 5 · 6 D 3 · 4 · 5 · 6 · 7 · 8 : 4 2.42/ 3 2.42/ 22.2 · 3/2 23.2 · 3 · 4/ Let Pn be the proposition that .2n/Š an D 2n.nŠ/2 ; D Section 1.2 Mathematical Induction 7 2 C 2 D ; 2 which is true for n D 1 (also for n D 2; 3; 4). If Pn is true then a D .2n C 1/.2n C 2/ a D .2n C 1/.2n C 2/ .2n/Š D .2n C 2/Š ; nC1 2.n C 1/2 2.n C 1/2 2n.nŠ/2 2nC1.n C 1/2 so Pn implies PnC1. (d) Rewrite the formula as anC1 D n C 1 n We compute a few terms to formulate Pn: a2 D 2a1 D 2; 3 2 3 2 32 4 3 4 3 32 43 a4 D 3 a3 D 3 2 D 3 · 2 : We take Pn to be the proposition that nn—1 an D .n — 1/Š; which is true for n D 1 (also for n D 2; 3; 4). If Pn is true, then n C 1 n n C 1 n nn—1 .n C 1/n anC1 D n so Pn implies PnC1. an D n .n — 1/Š D nŠ ; 1:2:10 . If an D nŠ then anC1 D .n C 1/a n D .n C 1/nŠ D .n C 1/Š, so Pn implies PnC1. However, Pn is false for all n, since an D 0 for all n, by induction. 1:2:11 . (a) If Pn is true then 1 C 2 C · · · C n C n C 1 D .n C 2/.n — 1/ C .n C 1/ so Pn implies PnC1. (b) No; see Example 1.2.1. n2 3n D 2 D .n C 3/n; 2 1:2:12 . Calculation shows that the first integer for which the inequality is true is n0 D 6. If n ≥ 6 and 1=nŠ > 8n=.2n/Š then 1 1 1 8n 1 .n C 1/Š D nŠ n C 1 > .2n/Š n C 1 : .A/ n 2 2 a3 D a2 D n : 8 Chapter 1 The Real Numbers n n n But 1 n C 1 8 — .2n C 2/.2n C 1/ 2n — 3 .n C 1/.2n C 1/ > 0; n ≥ 2: This and (A) imply that 1 8n 8 8nC1 .n C 1/Š > .2n/Š .2n C 2/.2n C 1/ D .2n C 2/Š ; so Pn implies PnC1. 1:2:13 . (a) If n ≥ 0, let Pn be the proposition that n D aq C r where q is an integer and 0 ≤ r < a. Then Pn is true if 0 ≤ n < a , since n D a · 0 C n. Pa is true, since a D a · 1 C 0 Now suppose that n ≥ a and P0, P1, . . . , Pn are true. Then n C 1 — a ≤ n, so n C 1 — a D q0a C r with 0 ≤ r < a, by the induction assumption. Hence n C 1 D .q0 C 1/a C r. This implies PnC1. (b) If b < 0 then 0 ≤ b C jbja D aq1 C r with 0 ≤ r < a, from (a); hence b D aq C r with q D q1 — jbj. (c) Suppose that aq C r D aq1 C r1, where q, q1, r and r1 are integers with (A) 0 ≤ r; r 1 < a. Then ajq — q1j D j r1 — rj. If q ¤ q1 then ajq — q1j ≥ a, which implies that jr1 — rj ≥ a, contradicting (A). Therefore q1 D q and r1 D r. 1:2:14 . (a) P1 is true, from the definition in Example 1.2.7. Now assume that Pk is true for some k ≥ 1. Suppose that that p divides p1p2 · · · pkpkC1 D ab, with a D p1 · · · pk and b D pkC1. From the given statement, p divides a or b. If p divides a then p D pi for some i in f1; : : : ; k g, by Pk . If p divides b then p D pkC1, by P1. Therefore Pk implies PkC1. (b) Let Pk be the proposition that the assertion is true if min .r; s/ D k. Then P1 is true, by the definition of prime in Example 1.2.7. Assume that Pk is true for some k ≥ 1. Now suppose that p1 · · · pkpkC1 D q1 · · · qs, where p1, . . . , pk, pkC1 and q1, . . . , qs are positive primes and s ≥ k C 1. Since pkC1 divides q1 · · · qs, (a) implies that pkC1 D qi for some i in f1; : : : ; s g. We may assume that prC1 D qs. Then p1 · · · pk D q1 · · · qs—1, and Pk implies that k D s — 1 and fp1; · · · ; pkg is a permutation of fq1; · · · ; qkg. This implies PkC1. 1:2:15 . P1 and P2 are true by inspection. Suppose that Pn and Pn—1 are true for some n ≥ 2. Then anC1 D Œ3n — .—2/n] C 6 3n—1 — .—2/n—1 D Œ3n C 6 · 3n—1] — Œ.—2/n C 6.—2/n—1] n D Œ3 C 2 · 3 ] — Œ.—2/ — 3.—2/ ] D 3nC1 — .—2/nC1; which implies PnC1. Apply Theorem 1.2.3. 1:2:16 . P1, P2 and P3 are true by inspection. Suppose that Pn, Pn—1 and Pn—2 are true for D Section 1.2 Mathematical Induction 9 p C p ; Z Z nC1 r dr D , verifying P . D p5 2 C 1 — p 5 2 C 1 D p5 — p 5 .1 — y/rC1 .1 — y/ C dy D .r C 2/.r C 3/ · · · .r C n C 2/ if r > —2. This — some n ≥ 3. Then anC1 D 9.3n—1 — 5n—1 C 2/ — 23.3n—2 — 5n—2 C 2/ C 15.3n—3 — 5n—3 C 2/ D .9 · 3n—1 — 23 · 3n—2 C 15 · 3n—3/ —.9 · 5n—1 — 23 · 5n—2 C 15 · 5n—3/ C .9 — 23 C 15/2 D .27 — 23 C 5/3n—2 — .45 — 23 C 3/5n—2 C 2 D 3n — 5n C 2; which implies PnC1. Use Theorem 1.2.3. 1:2:17 . Routine computations verify P1 and P2. Now suppose that n ≥ 2 and assume that .1 C p 5/k — .1 — p 5/k for 1 ≤ k ≤ n. Then Fk D 2k p5 .A/ FnC1 D Fn C Fn—1 D .1 C p 5/n — .1 — p 5/n 2n 5 .1 C p 5/n—1 — .1 — p 5/n—1 2n—1p5 1 1 C p 5 !n—1 " 1 C p 5 # 1 1 — p 5 !n—1 " 1 — p 5 # 1 1 C p 5 !n—1 " 3 C p 5 # 1 1 — p 5 !n—1 " 3 — p 5 # .1 C p 5/nC1 — .1 — p 5/nC1 D 2nC1 5 since p p 2 3 ˙ 5 D .1 ˙ 5/ : 1:2:18 . Let Pn be the proposition that the statement is true for all r > —1. Since 1 .1 y/r 0 1 dy D r C 1 , P0 is true. Now suppose that n ≥ 0 and Pn is true. Integra - tion by parts yields 1 y .1 — y/ dy D — ynC1 r C 1ˇ n C 1 Z 1 yn.1 — y/rC1 dy 0 .1 —Zy/r C1 0 r C 1 0 .A/ Z 1 n r 1 nŠ and (A) imply that 1 ynC1 0 .1 — y/r .n C 1/Š .r C 1/.r C 2/ · · · .r C n C 2/ nC1 2 2 2 2 1 n 0 0 1 D Z 2 2 2 4 n C 1 r C 1 y Pn implies that y C dy: 10 Chapter 1 The Real Numbers X ! n m n n n X X " ! This verifies the assertions in PnC1 for 1 ≤ m ≤! n, a nd we h!ave seen they hold automati- C ! n 1:2:19 . (a) nC1 n C 1 m mD0 tm D .1 C t/nC1 D .1 C t/.1 C t/n D .1 C t/ mX D0 n ! tm D mX D0 n ! tm C mX D0 n ! tmC1 D mX D0 n ! tm nC1 n tm mD1 m — 1 nC1 n D m C mD0 m — 1!# tm; since n ! D n ! D 0. Comparing coefficients of tm in the first and last expres - —1 sions shows that n C 1 n C 1! D n ! C n ! : .A/ m Now let Pn be the proposition that m m — 1 n ! D nŠ ; 0 ≤ m ≤ n: m mŠ.n — m/Š Then P0 is true, since .1 C t/0 D 1. Now suppose that Pn is true for some n ≥ 0. If 1 ≤ m ≤ n, then (A) and Pn imply that n C 1! D nŠ C nŠ m mŠ.n — m/Š .m — 1/Š.n — m C 1/Š D nŠ 1 C 1 .m — 1/Š.n — m/Š m n — m C 1 D nŠ n C 1 D .n C 1/Š : .m — 1/Š.n — m/Š m.n — m C 1/ mŠ.n — m C 1/Š cally for m D 0 and m D n C 1 (that is, PnC1. n C 1 0 X D n ! n C 1 n C 1 D 1). Therefore Pn implies (b) Set t D —1 in the identity .1 C t/n D obtain the second. mD0 tn to obtain the first sum, t 1 to m m m m n D Section 1.2 Mathematical Induction 11 2 D D D j D1 0 @ n X — (c) .x C y/n D yn.1 C x=y/n. From (a) with t D x=y, we see from this that .x C y/n D yn mD0 n ! .x=y/m D mX D0 n ! xmyn—m . 1:2:20 . We take constants of integration to be zero. Integrating by parts yields the first antiderivative A1.x/ D Z log x dx D x log x — Z 1 dx D x log x — x: Integration by parts yields the second antiderivative A2.x/ D Z .x log x — x/ dx D x2 2 log x — Z x C x dx D x2 3x2 : 2 4 We conjecture that an n-th antiderivative is xn An.x/ D nŠ .log x — kn/ ; .A/ for a suitable constant kn to be determined later. We have already verified (A) for n 1 and 2, with k1 1 and k2 3=2. If we assume that (A) is true for a given n then xnC1 1 Z xn n xnC1 AnC1 D .n C 1/Š log x — nŠ where n C 1 C knx dx D .n C 1/Š .log x C knC1/ ; 1 knC1 D kn C n C 1 : Now an easy induction yields kn D Pn 1=j . Therefore An D xn nŠ ln x — jX D1 1 1 A is an n-th antiderivative of log x for every n. 1:2:21 . f2.x1; x2/ D x1 C x2 C jx1 — x2j D 2 max .x1; x2/ from Exercise 1.1.1 (a) with a D x1 and b D x2. Now f3.x1; x2; x3/ D f2.x1; x2/ C 2x3 C jf2.x1; x2/ — 2x3j D 2 Œmax.x1; x2/ C x3 C j max.x1; x2/ — x3j] D 4 max .max .x1; x2/; x3/ from Exercise 1.1.1 (a) with a D max .x1; x2/ and and b D x3. Since max .max .x1; x2/; x3/ D max .x1; x2; x3/; f3.x1; x2; x3/ D 4 max .x1; x2; x3/: m m j n n 12 Chapter 1 The Real Numbers ˇ ˇ D n D I 2 sin x Now let Pn be the proposition that fn.x1; x2; : : : ; xn/ D 2n—1 max .x1; x2; : : : ; xn/: If Pn is true, then fnC1.x1; x2; : : : ; xnC1/ D fn.x1; x2; : : : ; xn/ C 2n—1xnC1 C fn.x1; x2; : : : ; xn/ — 2n—1xnC1 D 2n—1 Œmax.x1; x2; : : : ; xn/ C xnC1 Cj max .x1; x2; : : : ; xn/ — xnC1j] D 2n—1 max .max .x1; x2; : : : ; xn/; xnC1/ from Exercise 1.1.1 (a) with a D max .x1; x2; : : : ; xn/ and b D xnC1. Since max .max .x1; x2; : : : ; xn/; xnC1/ D max .x1; x2; : : : ; xnC1/ this implies PnC1. A similar argument using Exercise 1.1.1 (b) shows that gn.x1; x2; : : : ; xn/ 2n—1 min.x1; x2; : : : ; xn/. 1:2:22 . P1 is the assertion that sin x 1 — cos 2x : 2 sin x To verify this we let A D B D x in the second given identity, to see that cos 2x D cos2 x — sin2 x: Therefore 1 — cos 2x 2 sin x D 1 — cos2 x C sin2 2 sin x 2 sin2 x 2 sin x D sin x: Now suppose that n ≥ 1 and Pn is true. Then sin x C sin 3x C · · · C sin.2n C 1/x D Œsin x C sin 3x C · · · C sin.2n — 1/x] C sin.2n C 1/x D 1 — cos 2nx C sin.2n C 1/x (from P / 2 sin x 1 — cos 2nx C 2 sin x sin.2n C 1/x 2 sin x that is, sin x C sin 3x C · · · C sin.2n C 1/x D 1 — cos 2nx C 2 sin x sin.2n C 1/x : .A/ D D Section 1.3 The Real Line 13 D C D 2 sin x 2 sin x 2 To handle the product of sines in the numerator let A .2n 1/x and B x in the given identities. This yields cos 2nx D cos.2n C 1/x cos x C sin.2n C 1/x sin x cos.2n C 2/x D cos.2n C 1/x cos x — sin.2n C 1/x sin x: Subtracting the first of these identities from the second yields 2 sin.2n C 1/x sin x D cos 2nx — cos.2n C 2/x; and substituting this into (A) yields sin x C sin 3x C · · · C sin.2n C 1/x D 1 — cos 2nx C cos 2nx — cos.2n C 2/x ; so which is PnC1. sin x C sin 3x C · · · C sin.2n C 1/x D 1 — cos.2n C 2/x ; 1:2:23 . Let Pn be the stated proposition. P1 is trivial. Suppose that n > 1 and Pn—1 is true. If `n D n, Pn—1 implies Pn. If `n D s < n, choose r so that `r D n, and define `0i D `i if i ¤ r and i ¤ n; s if i D r; Then ˆ:n if i D n: Q.`1; `2; : : : ; `n/ — Q.`10 ; `20 ; : : : ; `n0 / D .xn — ys/2 C .xr — yn/2 — .xn — yn/ — .xr — ys/ Since `0n D n, Pn—1 implies that D 2.x n — xr /.yn — ys/ ≥ 0: (A) This and (A) imply Pn. 1.3 THE REAL LINE Q.`01; `02; : : : ; `n0 / ≥ Q.1; 2; : : : ; n/: 1:3:3 . Suppose that a 2 A. We consider two cases: (i) Suppose that a 2 X. Then a 2 A \ X and (b) implies that a 2 B \ X, which implies that a 2 B. (ii) Suppose that a 62 X. Since a 2 A [ X, (a) implies that a 2 B [ X. Therefore, a 2 B. Since a 2 X and a 62 X are the only two possibilities, it follows that a 2 B. Therefore, A ⊆ B. Similarly, B ⊆ A, so A D B. 1:3:6 . x 2 .S \ T /c , x 2 Sc or x 2 T c , x 2 Sc [ T c. 2 14 Chapter 1 The Real Numbers c c ˚ c ˇ 2 , 2 , 2 [ 2 T D ˇ D 2 2 ˚ ˇ ⊂ ¤ ; ¤ D — C 2 D — C ⊂ 1:3:8 . (a) If S1, . . . , Sn are open and x is in \n c i D1 cise 1.3.7). Since Ti is open, so is T .x0 — ‹; x0 C ‹/ ⊂ Ui . Hence, n i D1 Ui is a neighborhood of x0. x 2 .S [ T /c , x 2 Sc and x 2 T c , x 2 Sc \ T c. 1:3:7 . (a) x I x F for some F in F x F F F . (b) x 2 U c , x 2 Fc for every F in F , x 2 \ ˚F c ˇ F 2 F . such that .x — ‹i ; x C ‹i / ⊂ Si . If ‹ i D1 D min.‹1; : : : ; ‹n/, then .x — ‹; x C ‹/ ⊂ n i D1 Si . (b) T1 n 1.1=n; —1=n/ D f0g. 1:3:9 . (a) Let T D Sn Ti , where T1, . . . , Tn are closed. Then T c D Tn T c (Exer - (b) S1 n 1Œ1=n; 1/ D .0; 1/. D 1:3:10 . (a) Since U is a neighborhood of x0, there is an ‹ > 0 such that .x0 — ‹; x0 C ‹/ ⊂ U . Since U ⊂ V , .x0 — ‹; x0 C ‹/ ⊂ V . Hence, V is a neighborhood of x0. (b) Since U1, U2, . . . , Un are neighborhoods of x0, there are positive numbers ‹1, ‹2, . . . , ‹n such that .xT0 — n ‹i ; x0 C ‹i / T⊂ Ui .1 ≤ i ≤ n/. If ‹ D min.‹1; : : : ; ‹n/, then 1:3:12 . If x0 is a limit point of S, then every neighborhood of x0 contains points of S other than x0. If every neighbo rhood of x0 also contains a point in Sc, then x0 @S. If there is a neighborhood of x0 that does not contain a point in Sc, then x0 S0. These are tho only possibilities. 1:3:13 . An isolated point x0 of S has a neighborhood V that contains no other points of S. Any neighborhood U of x0 contains V \U , also a neighborhood of x0 (Exercise 1.3.10 (b)), so Sc \ U ¤ ;. Since x0 2 S \ U , x0 2 @S. 1:3:14. (a) If x0 2 @S and U is a neighbo rhood of x0 then, (A) U \ S ¤ ;. If x0 is not a limit point of S, then (B) U \ .S — fx0g/ D ; for some U . Now (A) and (B) imply that x0 2 S, and (B) implies that x0 is an isolated point of S. (b) If S is closed, Corollary 1.3.6 and (a) imply that @S ⊂ S; hence, S D S [ @S D S. If S D S, then @S ⊂ S. Since S0 ⊂ S, S is closed, by Exercise 1.3.12 and Corollary 1.3.6. 1:3:15 . False: 0 is a limit point of 1 n 1; 2; : : : , which consists entirely of isolated n points. 1:3:16 . (a) If ‹ > 0, then S \ .ˇ — ‹; ˇ/ ¤ ; and Sc \ .ˇ; ˇ C ‹/ ¤ ;. (b) If S is bounded below and ˛ D inf S then ˛ 2 @S. 1:3:17. inf S and sup S are in @S (Exercise 1.3.16) and @S S if S is closed (Exer- cise 1.3.14). 1:3:18 . Suppose that a S and H r r > 0 and .a r; a r/ S . Since S is open, H . If S R then H is bounded. Let p sup H . Then a p and a p are limit points of S and therefore in S, which is closed. Since S is open, there is an ‹ > 0 such that .a — p — ‹; a — p C ‹/ and .a C p — ‹; a C p C ‹/ are in S. Since .a — p C ‹=2; a C p — ‹=2/ ⊂ S it follows that .a — p — ‹; a C p C ‹/ ⊂ S, which contradicts the definition of p. Hence, S D R. 1:3:19. (a) If x0 is a limit point of @S and ‹ > 0, there is an x1 in .x0 — ‹; x0 C ‹/ \ @S. Since .x0 — ‹; x0 C ‹/ is a neighbo rhood of x1 and x1 2 @S, .x0 — ‹; x0 C ‹/ \ S ¤ ; and Si, there are positive numbers ‹1, . . . , ‹n c i D1 (Exercise 1.3.8). Hence, T is closed. i i D1 Section 1.3 The Real Line 15 2 — C ⊂ — C ⊂ 2 — C ⊂ — C ⊂ j D1 S D 1 H 2 D .s — js — x0j=2; s C js ˚— x0j=2/ ˇˇs 2 S is an open covering for S, but if s1, s2, n 2 1 2 1 2 1 2 .x0 — ‹; x0 C ‹/ \ Sc ¤ ;. Therefore, x0 2 @S and @S is closed (Corollary 1.3.6). (b) If x0 S0 then .x0 ‹; x0 ‹/ S for some ‹ > 0. Since .x0 ‹; x0 ‹/ S0 (Example 1.3.4), S0 is open. (c) Apply (b) to Sc. (d) If x0 is not a limit point of S, there is a neighborhood of x0 that contains no points of S distinct from x0. Therefore, the set of points that are not limit points of S is open, and the set of limit points of S is consequently closed. (e) S c D exterior of S, which is open, by (b). Hence, S is closed and S D S from (Exercise 1.3.14 (b)), applied to S. 1:3:19. (a) If x0 is a limit point of @S and ‹ > 0, there is an x1 in .x0 — ‹; x0 C ‹/ \ @S. Since .x0 — ‹; x0 C ‹/ is a neighbo rhood of x1 and x1 2 @S, .x0 — ‹; x0 C ‹/ \ S ¤ ; and .x0 — ‹; x0 C ‹/ \ Sc ¤ ;. Therefore, x0 is in @S and @S is closed (Corollary 1.3.6). (b) If x0 S0, then .x0 ‹; x0 ‹/ S for some ‹ > 0. Since .x0 ‹; x0 ‹/ S0 (Example 1.3.4), S0 is open. (c) Apply (b) to Sc. (d) If x0 is not a limit point of S, there is a neighborhood of x0 that contains no points of S distinct from x0. Therefore, the set of points that are not limit points of S is open, and the set of limit points of S is consequently closed. (e) S c D exterior of S, whic h is open, by (b). Hence, S is closed and S D S from (Exercise 1.3.14 (b)), applied to S. 1:3:21 . Since H 1 D ˚ .s — 1; s C 1/ ˇ s 2 S is an open covering for S, S ⊂ Sn .sj — 1; sj C 1/ for some s1 < s 2 < · · · < s n in S. Therefore, S ⊂ .s1 — 1; sn C 1/, and S is bo˚unded. If S is not closed, it hasˇ a limi t point x0 62 S (Theorem 1.3.5). Then . . . , sn are in S and 2ı D min jsi — x0j ˇ 1 ≤ i ≤ n , then i D1.si — jx0 — sij=2; si C jx0 — si j=2/ does not intersect .x0 — ı; x0 C ı/, which contains points of S. Therefore, no finite subcollection of H 2 covers S. 1:3:22 . (a) If t0 2 T and every neighborhood of t0 contains a point of S, then either t0 2 S or t0 is a limit point of S. In either case t0 2 S. Consequently, T ⊂ S. Conversely, if T ⊂ S D S [ @S, any t0 in T is in S or is a limit point of S (Exercise 1.3.14(a)); in either case, every neighborhood of t0 intersects S. (b) (a) with T R (reals) implies that S is dense in R if and only if every real number is in S or is a limit point of S. This is equivalent to saying that every interval contains a point of S. 1:3:23 . (a) x 2 .S1 \ S2/0 , x has a neighborhood N ⊂ S1 \ S2 , x 2 S0 and x 2 S0 , x 2 S0 \ S0. (b) x 2 S0 [ S0 ) x 2 S0 or x 2 S0 ) x has a neighborhood N such that N ⊂ S1 or N ⊂ S2 ) x has a neighbo rhood N ⊂ S1[S2 ) x 2 .S1[S2/0. 1:3:24. (a) x 2 @.S1 [ S2/ ) every neighbo rhood of x contains a point in .S1 [ S2/c and a point in S1 [ S2. If every neighborhood of x contains points in S1 \ S2, then x 2 @S1 \ @S2 ⊂ @S1 [ @S2. Now suppo se that x has a neighbo rhood N such that 16 Chapter 1 The Real Numbers 2 c N \ S1 D ;. If U is any neighborhood of x, then so is N \ U , and N \ U \ S2 ¤ ;, since N \ U must intersect S1 [ S2. This means that x 2 @S2 ⊂ @S1 [ @S2. A similar argument applies if x has a neighborhood N such that N \ S2 D ;. (b) x 2 @.S1 \ S2/ ) every neighbo rhood of x contains a point in .S1 \ S2/c and a point in S1 \ S2. If every neighborhood of x contains a point in .S1 [ S2/c, then x 2 @S1 \ @S2 ⊂ @S1 [ @S2. Now suppo se that x has a neighbo rhood N such that N ⊂ S1. If U is any neighborhood of x, then so is N \ U , and N \ U \ Sc ¤ ;, since N \ U must intersect .S1 \ S2/c. This means that x 2 @S2 ⊂ @S1 [ @S2. A similar argument applies if x has a neighborhood N such that N ⊂ S2. (c) If x 2 @S, then any neighbo rhood N of x contains points x0 in S and x1 not in S. Either x0 2 S or x0 2 @S. In either case N \ S ¤ ;. Since x1 2 N \ Sc, it follows that x 2 @S; hence, @S ⊂ @S. (d) Obvious from the definition of @S. (e) @.S — T / D @.S \ T c/ (definition of S — T ) ⊂ @S [ @T (Exercise 1.3,24(b)) D @S [ @T (Exercise 1.3.24(d)): C C D 2 — 2 2 2 2 2 CHAPTER 2 Differential Calculus of Functions of One Variable 2.1 FUNCTIONS AND LIMITS 2:1:1 . (a) If jxj ≤ 1, infinitely many y’s satisfy sin y D x; if jxj > 1, no y satisfies sin y D x. (b) ey > 0 for all y, while —jxj ≤ 0 for all x. (c) 1 C x C y > 0 for all x and y. (d) y.y — 1/ D x has two solutions for every x. 2:1:4 . (a) lim x2 2x 1 4; x!1 jx C 2x C 1 — 4j D jx C 2x — 3j D jx — 1jjx C 3j ≤ jx — 1j.jx — 1j C 4/ ≤ ı.ı C 4/ if jx — 1j < ı. Given ‹ > 0 choose ı ≤ min.1; ‹=5/. Then jx C 2x — 4j < .‹=5/.1 C 4/ D ‹ if jx — 1j < ı. x3 8 (b) lim D 10. Proof x!2 x — 2 ˇx2 — 8 ˇ 2 2 ˇ x — 2 — 10ˇ D jx C 2x C 4 — 10j D jx C 2x — 6j 17 D jx — 2jjx C 4j ≤ jx — 2j.jx — 2j C 6/ < ı.ı C 6/ 18 Chapter 2 Differential Calculus of Functions of One Variable ˇ ˇ ˇ C C — C D — ˇ ˇ x — 2 x!x0 1=2, so ˇ 1 C 1ˇ < 2x2. Therefore ˇ 1 C 1ˇ < ‹ if jxj < ı D min.1=p 2; p ‹=2/. —12 < 2 C 7 < — 8 ; ˇ2x C 3 ˇ < 12: x2 if jx — 2j < ı. Given ‹ > 0 choose ı ≤ min.1; ‹=7/. Then ˇ x2 — 8 ˇ ˇ x — 2 — 10ˇ < .‹=7/.1 C 6/ D ‹ if jx — 1j < ı: 1 1 ˇ ˇ ˇ p 2 (c) lim x!0 x2 — 1 D —1. Proof ˇ ˇx2 — 1ˇ C 1ˇ D ˇ x2 — 1 ˇ. If jxj < 1= 2 then 1 — x > ˇx2 — 1 ˇ ˇx2 — 1 ˇ (d) lim px D 2. Proof jpx — 2 ˇ x — 4 ˇ < jx — 4j , so jpx — 2j < ‹ if jx — 4j < x!4 ı D 2‹. (e) Since x3 — 1 D x2 C x C 1, j D ˇpx C 2 ˇ 2 f.x/ D x2 x 1 x — 2 C x D x2 C x C 1 C x.x — 2/ x — 2 2x2 x 1 x — 2 if x ¤ 1: We show that lim f.x/ 2. x!1 2x2 — x C 1 2x2 — x C 1 C 2.x — 2/ 2x2 C x — 3 .x — 1/.2x C 3/ f.x/C2 D x — 2 C2 D x — 2 D x — 2 D x — 2 if x ¤ 2: Therefore 2x C 3 jf .x/ C 2j D jx — 1j ˇ ˇ if x ¤ 1: .A/ To handle the multiplier 2x C 3 D 2.x — 2/ C 7 D 2 C 7 ; x — 2 x — 2 x — 2 we first restrict x to the interval .1=2; 3=2/ (so jx — 1j < 1=2). On this interval x — 2 3 ˇ x — 2 ˇ From this and (A), jf .x/ C 2j < 12jx — 1j if 0 < jx — 1j < 1=2. If ‹ > 0 is given, let ı D min.‹=12; 1=2/ . Then jf .x/ C 2j < ‹ if 0 < jx — 1j < ı. 2:1:5 . If lim x!x0 f.x/ D L according to Definition 2.1.2 and ‹0 > 0, there is a ı > 0 such that jf .x/ — Lj < ‹0 if 0 < jx — x0j < ı. Let ‹0 D K‹ to see that lim f.x/ D L according to the modified definition. If lim x!x0 f.x/ D L according to the modified definition and ‹0 > 0, there is a ı > 0 such that jf .x/ — Lj < K‹0 if 0 < jx — x0j < ı. Let ‹0 D ‹=K to see that lim x!x0 f.x/ D L according to Definition 2.1.2. D Section 2.1 Functions and Limits 19 1 1 ˇ — — D x C — !—2 x!x0— — x 2 if x > 0: then ˇ x C jxj — 0ˇ D 0 < ‹ if —ı < x < 0 and ˇx C jxj — 2ˇ D 0 < ‹ if 0 < x < ı. jx — 1j D jx — 1j 2:1:7 . (a) If x ¤ 0 then x C jxj D 0 if x < 0; If ‹ and ı are arbitrary positive numbers, ˇ x ˇ ˇ x ˇ x C jxj Therefore lim D 0 and lim x C jxj D 2. x!0— x 1 x!0C x 1 1 1 (b) Let f.x/ D x cos x C sin ˇ x C sin ˇ jxj . We first observe that lim x cos x!0 x D 0, since if jxj < ı ≤ ‹ then ˇx cos ˇ ≤ jxj < ‹. If x < 0 then f.x/ D x cos (since sin u D — sin.—u/), so xliˇ m x 1 f.x/ D 0. If x > 0 then f.x/ D x cos x 1 C 2 sin . !0— x 1 x Therefore lim x!0C f.x/ D 0 does not exist, because if it did we would have xlim 1 2 sin x lim x!0C f.x/ lim x cos x!0 x lim !0C f.x/ 0 lim !0C f.x/, which is impossible since 1 2 sin x oscillates between ˙2 as x ! 0C. (c) 8ˆ< 1 ; x > 1 x C x — 2 .x C 2/.x — 1/ ˆ:— x C 2 ; x < 1: lim D , since ˇ ˇ ˇ x — 1 ˇ < jx — 1j < ‹ if jx — 1j < min.1; 6‹/. x!1 x C 2 3 ˇ x C 2 — 3 ˇ D ˇ 3.x C 2/ ˇ 6 lim jx — 1j D 1 and lim jx — 1j 1 D — : x!1C x2 C x — 2 3 x!1— x2 C x — 2 3 (d) f.x/ D x2 x 2 px C 2 D p x C 2 .x — 1/ is undefined if x < —2, so x lim — f.x/ p p !—2 pdo es not exist. If x > —2 then jf .x/j D x C 2 jx — 1j ≤ x C 2jx C 2 — 3j ≤ x C 2 .jx C 2j C 3/. Therefore if ı ≤ min.1; ‹2=4/ then jf .x/j < ‹ if —2 < x < —2 C ı. Hence x lim C f.x/ D 0. 2:1:8 . If lim x!x0— h.x/ D —p .p > 0/, let ‹ D p=2 in Definition 2.1.5 (a). Then there is a ı > 0 such that jh.x/ — .—p/j < p=2 if x0 — ı < x < x0. This is impossible since jh.x/ — .—p/j D h.x/ C p ≥ p if h.x/ ≥ 0. 2:1:9 . (a) Letting ‹ D 1 in Definition 2.1.2, we see that there is a p > 0 such that jf .x/ — Lj < 1, so jf .x/j < jLj C 1, if 0 < jx — x0j < p. (b) For “ lim ” the set is of the form .x0 p; x0/. For “ lim x!x0C ” the set is of the form .x0; x0 C p/. 2 D .A/ 1 1 1 1 This and (A) imply that !0C D x x C 1 2 D 18 Chapter 2 Differential Calculus of Functions of One Variable D 2:1:10. (a) Let Pn be the proposition that lim Œf .x/]n Ln. P1 is true by assump- x!x0 Section 2.1 Functions and Limits 21 X n—1 ˇp ˇ ı p ı ˇ ˇ x!1 p D D C j — j !x0 x!1 jxj˛ sin x ˇ x˛ ˇ sin x ‹1=˛ tion. If n ≥ 1 and Pn is true then lim Œf .x/]nC1 D lim Œf .x/]n lim f.x/ (by x!x0 Eqn. (2.1.12)) D LnL (by Pn)D LnC1, which proves Pn. n—1 x!x0 x!x0 (b) In the identity un — vn D .u — v/ ur vn—1—r , let u D f.x/ and v D L to obtain r D0 (A) Œf .x/]n — Ln D .f .x/ — L/ X Œf.x/]r Ln—1—r : From Exercise 2.1.9, there is a p > 0 r D0 n such that jf .x/j < jLj C 1 if 0 < jx — x0j < p; hence (A) implies that jŒf .x/] — L j < Kjf.x/ — Lj if 0 < jx — x0j < p, where K is a constant. If ‹ > 0 there is a ı1 > 0 such that jf .x/ — Lj < ‹ if 0 < jx — x0j < ı1. Therefore jŒf .x/]n — Lnj < K‹ if 0 < jx — x0j < min.p; ı1/. 2:1:11 . Write f.x/ — pL D jf .x/ — Lj f.x/ C pL ≤ jf .x/ — Lj pL and apply Definition 2.1.2. 2:1:12 . Use Definitions 2.1.2 and 2.1.5. 2:1:14 . lim x f.x/ D L if f is defined on an interval .—1; b/ and for each ‹ > 0 there is a numb! er—˛1 such that jf .x/ — Lj < ‹ if x < ˛. 1 1 p 2:1:15 . (a) lim D 0 because < ‹ if x > 1= ‹. x!1 x2 C 1 x2 C 1 (b) lim sin x sin x 1 D 0 if ˛ < 0 because ≤ 1 < ‹ if x > . (c) lim x!1 x˛ does not exist if ˛ ≤ 0 since, for example, x˛ assumes the values ˙1 on very interval Œa; 1/. (d) lim e—x sin x D 0 because je—x sin xj D e—x < ‹ if x > — ln ‹. (e) lim x!1 tan x does not exist because tan x is not defined on Œa; 1/ for any a. (f) lim e—x2 e2x 0 because if ‹ < e then e—x2 e2x ee—.x—1/2 < ‹ if x > 1 x!1 ln.e=‹/ . 2:1:16 . For “ lim " statements such as “there is a ı > 0 such that f.x/ L < ‹ if x!1 0 < jx — x0j < ı” would be changed to “there is an ˛ such that jf .x/ — Lj < ‹ if x > ˛,”; for “ lim x!—1 " the last inequality would be replaced by “x < ˛.” 2:1:19. (a) x lim — f.x/ D —1 if f is defined on an interval .a; x0/ and for each real number M there is a ı > 0 such that f.x/ < M if x0 — ı < x < x0. (b) lim x!x0C f.x/ D 1 if f is defined on an interval .x0; b/ and for every real number M there is a ı > 0 such that f.x/ > M if x0 < x < x0 C ı. (c) lim x!x0C f.x/ D —1 if f is defined on an interval .x0; b/ and for every real number M there is a ı > 0 such that f.x/ < M if x0 < x < x0 C ı. x˛ n 20 Chapter 2 Differential Calculus of Functions of One Variable x!1 x!1 D 1 D 1 x!1 x!x0 x!x0 x!x0 x!x0 2:1:21 . (a) lim x!x0 f.x/ D 1 if f is defined on a deleted neighborhood of x0 and, for every real number M , there is a ı > 0 such that f.x/ > M if 0 < jx — x0j < ı. (b) lim x!x0 f.x/ D 1 if f is defined on a deleted neighborhood of x0 and for, every real number M , there is a ı > 0 such that f.x/ < M if 0 < jx — x0j < ı. 2:1:23 . (a) lim x!1 f.x/ D 1 if f is defined on an interval .a; 1/ and, for every real number M , there is an ˛ such that f.x/ > M if x > ˛. (b) lim x f.x/ D —1 if f is defined on an interval .—1; b/ and for, every real number M , th! e— re1 is an ˛ such that f.x/ < M if x < ˛. 2:1:25 . Suppose that L is finite and ‹ > 0. Since lim x!1 f.x/ D L, there is an ˛ 2 .a; 1/ such that jf .x/ — Lj < ‹ if x > ˛. Since lim g.x/ D 1, there is a 2 .c; 1/ such that g.x/ > ˛ if x > . Therefore jf .g.x// — Lj < ‹ if x > , so lim f.g.x// D L. Now suppose that L and M is arbitrary. Since lim x!1 f.x/ D 1, there is an ˛ 2 .a; 1/ such that f.x/ > M if x > ˛. Since lim g.x/ D 1, there is a 2 .c; 1/ such that g.x/ > ˛ if x > . Therefore f.g.x// > M if x > , so lim f.g.x// . x!1 The proof where L D —1 is similar to this. 2:1:26 . (a) Suppose that lim x!x0 f.x/ D L, and choose ı > 0 so jf .x/ — Lj < ‹0=2 if 0 < jx — x0j < ı. If x1 and x2 are in .x0 — ı; x0 C ı/ and distinct from x0, then jf .x1/ — f.x2/j ≤ jf .x1/ — Lj C jL — f.x2/j < ‹0, a contradiction. Hence lim f.x/ does not exist. (b) Replace “every deleted neighborhood of x0” by “every interval .x0; b/,” “every interval .a; x0/,” “every interval .a; 1/”, and “every interval .—1; b/”, respectively. 2:1:28. Let M be arbitrary. If —1 < L1 < L2 D 1, choose ı > 0 so that f.x/ > L1 — 1 and g.x/ > M — L2 C 1 if 0 < jx — x0j < ı; if L1 D L2 D 1, choose ı > 0 so that f.x/ > M=2 and g.x/ > M=2 if 0 < jx — x0j < ı. Then .f C g/.x/ > M if 0 < jx — x0j < ı, so lim .f C g/.x/ D 1 D L1 C L2. If —1 D L1 < L2 < 1, choose ı > 0 so that g.x/ < L2 C 1 and f.x/ < M — L2 — 1 if 0 < jx — x0j < ı; if L1 D L2 D —1 choose ı > 0 so that f.x/ < M=2 and g.x/ < M=2 if 0 < jx — x0j < ı. Then .f C g/.x/ < M if 0 < jx — x0j < ı, so lim .f C g/.x/ D —1 D L1 C L2. 2:1:29 . Suppose that L1 D 1, 0 < p < L2 ≤ 1, and M > 0. Choose so g.x/ > p and f.x/ > M=p if x > ; then .fg/.x/ > M if x > , so lim .fg/.x/ D 1 D L1 · L2. Similar arguments apply to the other cases. x!1 2:1:30 . (a) Let M be arbitrary. Suppose that L1 D 1. If 0 < L2 < 1 choose ı > 0 so 0 < g.x/ < 3L=2 and f.x/ > 2M=3L 2 if 0 < jx — x0j < ı. Then .f=g/.x/ > M if 0 < jx — x0j < ı,so lim .f=g/.x/ D 1 D L1=L2. Similar arguments apply to the other cases where L1 D ˙1 and 0 < jL2j < 1.