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SalasSV_10_04_ex_ans

# SalasSV_10_04_ex_ans - ANSWERS TO ODD-NUMBERED EXERCISES 27...

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Unformatted text preview: ANSWERS TO ODD-NUMBERED EXERCISES 27. converges to 35. b < 1 2 A-73 29. converges to e2 31. diverges 33. (a) 0 (b) π 4 (c) 1 2 √ √ √ nn a + bn = b n (a/b)n + 1 < b n 2. Since 21/n → 1 as n → ∞, it follows that n an + bn → b by the pinching theorem. 39. 1+ 1 n n +1 37. Use |(an + bn ) − (L + M )| ≤ |an − L| + |bn − M |. = 1+ 1 n n 1+ 1 1 . Note that 1 + n n n → e and 1+ 1 n → 1. 41. Imitate the proof given for the nondecreasing case in Theorem 10. 3. 6. 43. Let > 0. Choose k so that, for n ≥ k , L − < cn < L + and an ≤ bn ≤ cn . L − < an < L + , For such n, L − < bn < L + . 45. Let > 0. Since an → L, there exists a positive integer N such that L − < an < L + for all n ≥ N . Now an ≤ M for all n, so L − < M , or L < M + . Since is arbitrary, L ≤ M . 47. Assume an → 0. Let > 0. There exists a positive integer N such that |an − 0| < Now assume |an | → 0. Since −|an | ≤ an ≤ |an |, an → 0 by the pinching theorem. 49. By the continuity of f , f (L) = f ( lim an ) = lim f (an ) = lim an+1 = L. n→∞ n→∞ n→∞ for all n ≥ N . Since ||an | − 0| ≤ |an − 0|, it follows that |an | → 0. 53. converges to 0 51. Use Theorem 10.3.12 with f (x) = x1/p . 63. L = 0, n = 7 55. converges to 0 57. diverges √ 3+ 5 67. (a) (b) 3 2 69. (a) a2 0.540302 a3 0.857553 59. L = 0, n = 32 61. L = 0, n = 4 65. L = 0, n = 65 a4 0.654290 a5 0.793480 a6 0.701369 a7 0.763960 a8 0.722102 a9 0.750418 a10 0.731404 (b) 0. 739085; it is the ﬁxed point of f (x) = cos x. SECTION 10.4 1. converges to 1 15. converges to 1 27. converges to e−1 3. converges to 0 17. converges to π 29. converges to 0 5. converges to 0 19. converges to 1 7. converges to 0 9. converges to 1 23. diverges 11. converges to 0 25. converges to 0 37. (a) 2 13. converges to 1 21. converges to 0 33. converges to e x 31. converges to 0 35. converges to 0 (b) 0 (c) 1 √ √ √ √ √ n+1− n √ 1 39. n + 1 − n = √ √ ( n + 1 + n) = √ √ →0 n+1+ n n+1+ n 41. (b) 2π r . As n → ∞, the perimeter of the polygon tends to the circumference of the circle. 47. (a) mn+1 − mn = 43. 1 2 45. 1 8 1 1 (a1 + · · · + an + an+1 ) − (a1 + · · · + an ) n+1 n 1 = nan+1 − (a1 + · · · + an ) > 0 since {an } is increasing. n(n + 1) |a1 + · · · + aj | |a1 + · · · + aj | n−j + → 0, and therefore for n sufﬁciently large (b) We begin with the hint mn < . Since j is ﬁxed, n 2 n n |a1 + · · · + aj | n−j < . Since < , we see that, for n sufﬁciently large, |mn | < . This shows that mn → 0. n 2 2 n 2 √ √ √ √ √ 49. (a) Let S be the set of positive integers n(n ≥ 2) for which the inequalities hold. Since ( b)2 − 2 ab + ( a)2 = ( b − a)2 > 0, it follows that √ √ a1 + b 1 a1 + b1 a+b > ab and a1 > b1 . Now a2 = < a1 and b2 = a1 b1 > b1 . Also, by the argument above, a2 = > a1 b1 = b2 , and so 2 2 2 ak + bk ak + a k ak + bk a1 > a2 > b2 > b1 . Thus 2 ∈ S . Assume that k ∈ S . Then ak +1 = < = ak , bk +1 = ak bk > b2 = bk , and ak +1 = > k 2 2 2 ak bk = bk +1 . Thus k + 1 ∈ S . Therefore the inequalities hold for all n ≥ 2. (b) {an } is a decreasing sequence which is bounded below. {bn } is an increasing sequence which is bounded above. Let La = lim an , Lb = lim bn . Then an = n→∞ n→∞ an−1 + bn−1 La + L b implies La = and La = Lb . 2 2 A-74 ANSWERS TO ODD-NUMBERED EXERCISES ∼ 51. The numerical work suggests L = 1. Justiﬁcation: Set f (x) = sin x − x2 . Note that f (0) = 0 and for x close to 0, f (x) = cos x − 2x > 0. Therefore sin x − x2 > 0 for x close to 0 and sin (1/n) − 1/n2 > 0 for n large. Thus, for n large, 1 1 1 < sin < n2 n n ↑ 1 n2 1 n1/n 1/n | sin x| ≤ |x| for all x 1/n 1 < sin n < sin 1 n < < 1 n 1 . n1/n 1/n 2 1/n As n → ∞ both bounds tend to 1 and therefore the middle term also tends to 1. 53. (a) a3 2 a4 3 a5 5 a6 8 a7 13 a8 21 a9 34 a10 55 (b) r1 1 r2 2 r3 1.5 r4 1.6667 r5 1.6000 r6 1.625 (c) L = √ 1+ 5 ∼ = 1. 618033989 2 SECTION 10.5 1. 0 3. 1 23. 1 3 5. 1 2 7. ln 2 27. 4 x→0 9. 29. 1 2 1 4 11. 2 31. 1 2 13. 33. 1 1+π 1−π 35. 1 15. 1 2 17. π 39. 1 e 2 19. − 1 2 41. 1 21. −2 √ 6 25. − 1 8 37. 0 47. − 43. lim (2 + x + sin x) = 0, lim (x3 + x − cos x) = 0 x→0 45. a = ± 4, b = 1 49. f (0) 51. (a) 1 (b) − 1 3 53. 3 4 55. (a) f (x) → ∞ as x → ± ∞ (b) 10 57. (b) ln 2 ∼ 0. 6931 = SECTION 10.6 1. ∞ 3. −1 27. e3 29. e 5. ∞ 31. 0 7. 1 5 9. 1 35. 0 11. 0 37. 1 13. ∞ 39. 1 15. 1 3 17. e 43. 1 19. 1 45. 0 21. 1 2 23. 0 25. 1 33. − 1 2 41. 0 47. y-axis vertical asymptote y 49. x-axis horizontal asymptote y 51. x-axis horizontal asymptote y y = x2 x (–1, – e ) 1 (2, 4 e –2) x x 53. √ b2 b x 2 − a2 + x x − a2 − x = √ a a x 2 − a2 + x x→0+ b −ab →0 ( x 2 − a2 − x ) = √ a x 2 − a2 + x as x → ∞ 55. example : f (x) = x2 + (x − 1)(x − 2) x3 57. lim cos x = 0 59. (a) Let S be the set of positive integers for which the statement is true. Since lim lnx = 0, 1 ∈ S . Assume that k ∈ S . By L opital’s rule, ’H ˆ x (since k ∈ S ). x→∞ x→∞ lim (ln x)k +1 ∗ (k + 1)(ln x)k = lim =0 x→∞ x x Thus k + 1 ∈ S , and S is the set of positive integers. (b) Choose any positive number α . Let k − 1 and k be positive integers such that k − 1 ≤ α ≤ k . Then, for x > e, (ln x)k −1 (ln x)α (ln x)k ≤ ≤ x x x and the result follows by the pinching theorem. ...
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