Business Calc Homework w answers_Part_72

Business Calc Homework w answers_Part_72 - 356 Chapter 8...

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Unformatted text preview: 356 Chapter 8 Review 1 9. lim x→1 x 1 ln x 1 ln x x 1 (x 1) ln x lim x→1 1 x lim x 1 x→1 x x→ ln x 1 2 ln x 19. lim x→ 0 10. The limit leads to the indeterminate form ln f (x) lim x→0 . f (x) x→ g(x) 20. lim ln (1 1/x) 1/x 1x x lim x→ 1 x x x→ csc 1 x 1/x x→ lim 1 /x 2 1 1/x lim x→ lim 1/x 2 x→0 x→0 lim eln f (x) e0 x→0 lim x2 1 lim x x 0 1 x lim x2 1 x2 x2 1 lim 1 1 x2 x→ x→ x→ 11. The limit leads to the indeterminate form 0 . x→0 since x and lim x 2 lim 1 1/x 2 1 1 f grows at the same rate as g. (tan ) ln f ( ) 1 tan x ln (1 1/x) 1/x 0 f( ) lim 1 1 x x ln 1 lim 1 1 f grows faster than g. 1x x x→0 1 1/x 2 1 ex x→ 100 lim f (x) g(x) lim tan 1 lim x→ 1/x f grows faster than g. x→ f (x) x x x/100 x x→ xe f (x) x→ g(x) x x ln x x/x lim x→ 18. lim 1 lim x→1 1 f (x) g(x) f grows at the same rate as g. 1 1 1 lim x→1 x 17. lim ln (tan ) 1/ f (x) x→ g(x) ln (tan ) 1/ ln (tan ) 21. lim lim x ln x lim x→ lim x (ln x)(1 2 x→0 (ln x)/ln 2 x→ 2 lim log2x lim xln x sec2 tan 1 x→0 xln x log x x→ x 2 lim sin x→ cos lim x→ 2 lim sin2 x→0 cos2 Note that 1 0 1/ln 2) 1 (ln x)(1/ln 2 x 1 ln 2 1) 0 0 since ln 2 1. f grows slower than g. ln f ( ) lim (tan ) lim e x→0 12. lim → 2 x 3 3x 2 2 x x→ 2x 1 3 3x 2 x 4 x3 x→ x 1 2 14. lim x→ f (x) g(x) lim x→ 1 f (x) x→ g(x) 22. lim sin t lim 2 t→0 t 13. lim 15. lim e x→0 1 sin 0 cos t lim 2t t→0 3x 2 4x x→ 6x 1 6x 3 x→ 4x 1 3x 2 lim lim x 5x lim x→ lim x→ 3 x→ 2 lim 6x 6 23. lim 4 x→ 6 12x 2 6x 0 lim x→ lim x→ log2 x log3 x f (x) x→ g(x) lim x→ lim lim x→ x→ lim f grows at the same rate as g. (ln x)/(ln 2) (ln x)/(ln 3) ln 3 ln 2 ln 2x ln x 2 lim x→ lim x→ 2x 3 ln x ln 2 2 ln x f grows at the same rate as g. 24. lim 1 5 f (x) g(x) x→ f (x) x→ g(x) 2x x x→ 3 lim x 0 since 2 3 1/x 2/x 1 2 f grows slower than g. f grows at the same rate as g. 16. lim x lim x→ 10x 3 2x 2 ex 30x2 4x ex 60x 4 ex 60 0 ex f grows slower than g. lim x→ 1. Chapter 8 Review f (x) x→ g(x) 25. lim 1 (1/x) 1/x tan lim x→ 1 1 lim 34. lim 2( 1 (1/x) ) 1 2 x→ 2 (x 2 x lim 2 sec 1 x 1 x→ 28. (a) lim f (x) sin x 1 ex x→0 (b) Define f (0) lim x→0 1 (ln 2)(cos x)2 ex ln 2. ln 2 39. x 9 ln x x→0 1/x 1/x lim 1/x 2 x→0 lim ( x) 2 2 3 cos x 2 0 sin 1 x2 lim 1 d 1 3 1 0 x 3 b dx 9 1 lim x 2 b→3 0 C dx x2 9 True 1 x2 30. lim x→ 1 x4 1 x4 lim lim (x 2 b→3 1) x→ lim False 31. lim x→ x b→3 x ln x 1 lim x→ 1 x 1 1 0 lim 1 ln x False 32. lim x→ ln (ln x) ln x lim 1 x ln x x→ 1 x True tan 1 x 1 x→ 33. lim True 2 2 2 C 0. x→ d, 3 cos 1 x2 1 d x→0 2 4 29. lim x 1 x x→ 1 3 cos dx lim 1 1 x 1 x 3 sin , dx x→0 (b) Define f (0) 0 1 True lim x ln x x→0 lim x→ 38. lim 2 lim lim True sin x x→0 0 1 (1/x)2 1 f grows faster than g. 27. (a) lim f (x) x2 1 x x→ ln 2x 37. lim x→ ln x 3 2x x→ 1 lim ) (1/x ) x→ 1 True 1 1 1 x2 ln x 36. lim 1 x→ x lim lim x→ 0 1 True sin 1 (1/x) (1/x 2) x→ f (x) x→ g(x) 1 x4 x2 x→ 1 x4 35. lim f grows at the same rate as g. 26. lim 1 lim 1 x4 1 x2 True 1 x→ x→ 2 x lim 2 x (1/x) x→ 1 x4 x→ 0 sin sin 1x 3 b 0 1b 3 2 2 357 358 Chapter 8 Review 40. u ln x dv ln x dx 43. 3 1 dx x du dx v x ln x x dx 1 x lim ln x dx b→0 1 x ln x b→0 b ln b 1 1 dy y 2/3 0 b→0 1 lim 3y 1/3 3 3x dy 3 b→ 0 where A 3 1 6 1 d 2 1 2 1)3/5 ( 2 1)3/5 1 B x 2 5 ( 2 lim 5 (b 2 b→ 1 0 1 0 d 1)3/5 ( lim b→ 1 lim b→ 1 lim b→ 1 0 2 d ( 1)3/5 5 2 1) 4B)x 2 1 dx x2 3x 4x 3 ln 3 C(4x (B 1, C 1) 4C)x C 1)3/5 ( b b→ b 1)2/5 ln 2 1) 5 2 2/5 5 2 du 5 ( 2 1)2/5 0 u x 1 dv dx 2x x e v x 2e b 5 2 5 (b 2 5 2 5 2 du 2 dx 0 1)2/5 x 2e x x x e v e x 2e 0 x dx 1 dx x x 2e x 2x e x x 2x e x b x 2e x lim x 2e x lim b2 eb lim b→ b→ b→ 1 dx x 2e dx 1 dx 2x e dv 3 4 b x e x ln 3 ln 1 dx x2 1 x ln x 1 b 1 ln 3 2x dx x 2e 1)3/5 4b x2 45. u d ( 1 4 1 1 b lim ln 1 x 4 4x 1 ln 4x b→ b 1 lim b→ d 1 1)3/5 ( lim b→ 1 1 3 C x2 Bx(4x 4, B d lim 1)3/5 b→ 1 ( 0 d ( b d ln 3x 1 dx x 2(4x 1) 1 lim 42. 2 b A (A 3 b 1 b b→0 2 lim 4x Ax 2 1 2 b x b 1 dx x2 3x 4x 3 3x 1 x 2(4x 1) 3b1/3] lim [3 3 1 3] 2/3 y lim 1 44. dy ln x x b→ b b→0 dy y 2/3 1 1 b→0 1 lim ln b lim [3b1/3 dx 2 3 b→ 1 3y1/3 1 x ln x b→ dy y 2/3 2/3 y 1 x 3 lim ln 1 lim b→0 b→ b lim ( b) b dy y 2/3 2A 1. lim b→0 b→0 0 B)x lim ln b 1/b 1/b lim 1/b 2 b→0 lim 1 (A b 2 dx x(x 2) lim 1 dy y 2/3 2 1, B b→0 1 1 0 b) 3 1 dy y 2/3 Bx B b b→0 41. 2) where A x lim ( 1 0 x b lim 1 3 A( x 2 1 ln x dx 0 b→ A x x(x C 2 dx x(x 2) lim 2) 2 x ln x b 2 dx 2x x 2 x 2e 2e dx x C dx 0 2x e 2b eb 2 eb x 2e x b 0 2 2 359 Chapter 8 Review 46. u x du e 3x dx dv dx 0 xe 1 3x e 3 v 1 x 3 1 x 3 x e3x dx dx has infinite discontinuities at t 1 3x e dx 3 e3x 1 3x e 9 3x e 0 3x Note that the integral must be broken up since the integrand lim xe b→ 0 b 3x lim b→0 dx lim b→ B t2 b→ C ln (t 1)2 t2 1 tan lim 1 b (b 1)2 ln 2 b 1 1 3b e 9 1 9 tan b→0 b 4t 3 t 1 47. 2 t (t 1)(t 2 1) t 1 A t t 1)(t 2 At(t 1) 22 Ct (t B E)t (t B E)t 3 D (A B)t 0 4 dx x 2 16 A D B 0, C A E B 0, D A E B 1x tan 2 0 lim 1. 0 b 4 dx x 2 16 Solving the system simultaneously yields 0, B 1, C 2, D 2, E 0 4 dt t2 2 dt t1 2t t dt t2 2 dt t1 4 dx x 2 16 2t dt t2 1 2 1 dt 1 1 ln t (t 1) t2 1 ln 1 tan 1 t C lim 2 1 t tan 1 t → C t1 4t dt t 2(t 1)(t 2 1) 1 t1 4t dt t 2(t 1)(t 2 1) 2 1 4t 3 t 1 dt t (t 1)(t 2 1) 2 1 1/2 3 t1 4t dt t 2(t 1)(t 2 1) 1 2 1 lim lim → g( ) d → 1 1 b→ 3 1 t1 4t dt t 2(t 1)(t 2 1) and 1 1 1 1 2 d 1 1 d b lim ln b→ 4t 3 t 1 dt t (t 1)(t 2 1) 1 lim ln b 2 b→ 3 2 1 2 . Both are positive continuous functions on [1, ). b 0 2 2 lim 3 1/2 f( ) g( ) Since 3 0 1 g( ) 2 ln t 4 49. Use the limit comparison test with f( ) 1 dt t2 1 2 2 0 1b lim tan t1 4t dt t 2(t 1)(t 2 1) 1 t 4 b→ 3 2 b 1x lim tan 1. b 1b b→ A 4 4 dx x 2 16 lim b→ 4 tan b→ B 4 dx x 2 16 0 1x tan b→ 4, 0 4 dx x 2 16 b lim 1 and 1 C using Formula 16 with a 4 4 dx lim 16 b→ x B Equating coefficients of like terms gives C 4 dx x 2 16 2 0 A 1 4 4 dx 16 x 48. 1) E) t 2 C ln 2 t 1) 2 (A 1 1 Since this limit diverges, the given integral diverges. 1)(t 2 (Dt D)t 4 (A E 1 B( t 1) C (A Dt t2 1 b 1 t b→0 b 1 3b be 3 4t 3 t 1 dt t (t 1)(t 2 1) 2 lim 0 1 3x e 9 1 9 lim 4t 3 1 b 1 3x xe 3 1. 4t 3 t 1 dt t 2(t 1)(t 2 1) 1 C 0 and t t1 4t dt t 2(t 1)(t 2 1) , we know that 1 g( ) d diverges and so 1 f( ) d diverges. This means that the given integral diverges. 360 Chapter 8 Review x 50. Evaluate e u cos x dx using integration by parts. cos x du e dv sin x dx x v cos x dx x Evaluate sin x e u cos x x sin x e ex dx dv cos x dx v x x x e x dx e 0 ex e dx x x lim b→ sin x e dx e sin x e cos x dx lim e x cos x dx x e cos x x e sin x e x b→ cos x dx 0 x 2e cos x dx x e sin x x e cos x dx ex e b→ C1 0 x u e x e cos x dx sin x e x cos x b cos u du lim e b→ 0 x 1 eb cos x dx lim x 2 e x cos x b 0 e sin x b cos b 2 b→ e lim b sin b 2 b→ 1 2 1 2 4 x eb 4 4 4 Since these two integrals converge, the given integral converges. 54. The integral has an infinite discontinuity at x Note that we cannot use a comparison test since e x cos x 0 for some values on [0, ). 1 dx x (1 e x ) ln z on [e, ) z b dz b lim lim ln z lim (ln b 1) e b→ b→ e z b→ 1 dz diverges, so the given Since this integral diverges, ez 2 1 4x 2 1 0 dx 4x 2 1 x 2(1 ex ) 1 dx 4x 2 lim b→0 b 1 11 4x b lim b→0 1 0 0. dx x 2(1 e x ) dx x 2(1 e x ) ex on (0, 1] since 1 Since this integral diverges, integral diverges. 1 dx x 2(1 e x ) 0 0 0 dx x (1 e x ) 2 1 1 51. 0 z dz z e C b 0 ex b→ ex x 1 0 e 1 lim tan C 2 dx e ex 1 0 b lim tan b→ e ex dx ex e lim x tan x tan 4 b C 0 dx e 1 tan b→ u x ex b lim x dx e ex 0 dx e 1 tan 0 dx e e dx 1) du u2 1 x e x dx (e x)2 1 x e , du x x e dx e (e 2 x x x Let u dx using integration by parts. sin x du x dx x e x e x e dx e e 53. 4 on (0, 1]. 1 4 lim b→0 1 4b dx diverges, so the x 2(1 e x ) given integral diverges. t e 52. 0 e t on [1, ) t e 1 t 55. x 2 b dt lim b→ e e lim e b→ dt 1 lim b→ t x2 tb 2x 1 b 1 e 7x 2x 1 7x 12 1 (x 4)(x A x 3) B 4 x 3 A( x 3) B(x 4) (A 1 e Since this integral converges, the given integral converges. 12 B)x 3A 4B Equating coefficients of like terms gives A B 2 and 3A 4B 1. Solving the system simultaneously yields A 2x 1 dx x 2 7x 12 9 ln x 9, B 7 dx x3 9 dx x4 4 7 ln x 3 C 7. ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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