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SalasSV_10_05_ex_ans

# SalasSV_10_05_ex_ans - A-74 ANSWERS TO ODD-NUMBERED...

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Unformatted text preview: 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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