SalasSV_10_06_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. ANSWERS TO ODD-NUMBERED EXERCISES 61. The limit has the form 1∞ ; use L opital’s rule ’H ˆ 63. (a) Ab = 1 − (1 + b) e−b (b) xb = b→∞ A-75 2 − (2 + 2b + b2 )e−b , 1 − (1 + b)e−b b→∞ yb = b→∞ 1 4 − 1 (1 + 2b + 2b2 )e−2b 4 2[1 − (1 + b)e−b ] 1 8 (c) lim Ab = 1; x→0+ lim xb = 2, lim yb = 65. lim (1 + x2 )1/x = 1. 67. lim g (x) = −5/3. x→∞ SECTION 10.7 1. 1 23. 4 3. 1 π 4 5. diverges 7. 6 9. 1 π 2 11. 2 31. 1 2 13. diverges 33. 2 e − 2 15. − 1 4 17. π 1 32 π 16 π 16 19. diverges 37. π 2 21. ln 2 39. π 25. diverges 27. diverges 29. diverges 35. (a) converges: (b) converges: (c) converges: (d) diverges −1 41. surface area = 1 ∞ 2π 1 x 1+ 1 dx = 2π x4 1 0 ∞ 1 √ x4 + 1 dx > 2π x3 2 1 0 ∞ 1 1 dx = ∞ x 43. (a) y (b) 2 (c) V = π 1 √ x dx = π 1 dx, diverges x 45. (a) y (b) 1 (c) 1 π 2 (d) 2π √ √ (e) π [ 2 + ln(1 + 2)] x 1 x 2 47. (a) The interval [0, 1] causes no problem. For x ≥ 1, e−x ≤ e−x and 1 ∞ e−x dx is finite. (b) Vy = 0 ∞ 2π x e−x dx = π ∞ 4 3 2 49. (a) y (b) 51. converges by comparison with 0 (c) 2π (d) 8 π 7 1 x dx x3/2 53. diverges since for x large the integrand is greater than ∞ 1 and x ∞ 1 b 0 1 dx diverges x =∞ ∞ 55. converges by comparison with 1 dx x3/2 57. (a) 0 2x dx = lim b→∞ 1 + x2 b b 0 2x dx = lim ln (1 + x2 ) b→∞ 1 + x2 b −b (b) lim b→∞ −b 2x dx = lim ln (1 + x2 ) b→∞ 1 + x2 = lim b→∞ ln (1 + b2 ) − ln (1 + b2 ) = lim 0 = 0 b→∞ √ 59. L = (a 1 + c2 / c) ecθ1 65. f (x) ≥ 0 for all x and 69. Observe that F (t ) = result follows. 1 t ∞ −∞ 61. 1 ; dom (F ) = (0, ∞) s ∞ 0 63. s ; dom (F ) = (0, ∞) s2 + 4 67. 1 k n 1 f (x) dx = 6x dx = 1 (1 + 3x2 )2 f (x) dx is continuous and increasing, that an = f (x) dx is increasing, and that an ≤ 1 t f (x) dx ≤ an+1 for t ∈ [n, n + 1] . The ...
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