For n 1 y x a so there is no point of inection 35

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Unformatted text preview: →0 x x→0 x→+∞ x→+∞ ln cos(2/x) (−2/x2 )(− tan(2/x)) − tan(2/x) = lim = lim 2 x→+∞ x→+∞ x→+∞ 1/x −2/x3 1/x 2 x→+∞ lim x→0 (2/x2 ) sec2 (2/x) = −2, lim y = e−2 x→+∞ −1/x2 1 1 − sin x x lim x→0 b ln(1 + a/x) ab = lim = ab, lim y = eab x→+∞ 1 + a/x x→+∞ 1/x ln(2 − x) 2 sin2 (πx/2) = lim = 2/π , lim y = e2/π x→1 cot(πx/2) x→1 x→1 π (2 − x) x→1 = limx→+∞ 31. x→0 3 ln(1 + 2x) 6 = lim − = −6, lim y = e−6 x→0 x→0 x 1 + 2x = lim x→0 x − sin x 1 − cos x sin x = lim = lim =0 x→0 x cos x + sin x x→0 2 cos x − x sin x x sin x 1 − cos 3x 3 sin 3x 9 9 = lim cos 3x = = lim x→0 x→0 2 x2 2x 2 33. (x2 + x) − x2 x lim √ = lim √ = lim 2+x+x 2+x+x x→+∞ x→+∞ x→+∞ x x 34. ex − 1 − x ex − 1 ex = lim x = lim x = 1/2 x−x x−1 x→0 xe x→0 xe + e x→0 xe + 2ex 35. lim [x − ln(x2 + 1)] = lim [ln ex − ln(x2 + 1)] = lim ln x→+∞ lim x→+∞ 38. (a) x→+∞ x x→+∞ ex , x2 + 1 x x 1 = lim ln = ln(1) = 0 1 + x x→+∞ 1/x + 1 ln x 1/x 1 = lim = lim =0 n x→+∞ nxn−1 x→+∞ nxn x xn nxn−1 = lim = lim nxn = +∞ lim x→+∞ ln x x→+∞ 1/x x→+∞ lim x→+∞ 0 3x2 − 2x + 1 because it is not a form x→1 3x2 − 2x 0 (a) L’Hˆpital’s Rule does not apply to the problem lim o (b) 40. = 1/2 e e e = lim = lim = +∞ so lim [x − ln(x2 + 1)] = +∞ x→+∞ x2 + 1 x→+∞ 2x x→+∞ 2 lim ln x→+∞ (b) 39. 1 + 1/x + 1 lim x 36. 1 3x2 − 2x + 1 =2 x→1 3x2 − 2x lim 4x3 − 12x2 + 12x − 4 12x2 − 24x + 12 24x − 24 = lim = lim =0 x→1 4x3 − 9x2 + 6x − 1 x→1 12x2 − 18x + 6 x→1 24x − 18 lim Exercise Set 4.7 41. lim x→+∞ 132 1/(x ln x) 2 √ = lim √ =0 x→+∞ 1/(2 x) x ln x 0.15 100 10000 0 42. y = xx , lim+ ln y = lim+ x→0 x→0 ln x = lim −x = 0, lim+ y = 1 x→0 1/x x→0+ 1 0 0.5 0 43. y = (sin x)3/ ln x , lim+ ln y = lim+ lim y = e3 x→0 x→0 x 3 ln sin x = lim+ (3 cos x) = 3, x→0 ln x sin x 25 x→0+ 0 0.5 19 44. lim − x→π/2 4 sec2 x 4 = lim =4 sec x tan x x→π/2− sin x 4 0.5 p/ 2 0 45. e−x ln x − 1 ; e−x e−x ln x 1/x = lim = 0 by L’Hˆpital’s Rule, so o lim e−x ln x = lim x→+∞ x→+∞ ex x→+∞ ex −x e ln x − 1 lim [ln x − ex ] = lim = −∞ x→+∞ x→+∞ e−x ln x − ex = ln x − 1 = 0 0 3 -16 46. lim [ln ex − ln(1 + 2ex )] = lim ln x→+∞ x→+∞ 1 1 = ln ; = lim ln −x x→+∞ e +2 2 horizontal asymptote y = − ln 2 ex 1 + 2ex -0.6 0 -1.2 12 133 47. Chapter 4 ln(ln x) 1 = lim = 0; x→+∞ x ln x x lim y = 1, y = 1 is the horizontal asymptote 1.02 y = (ln x)1/x , lim ln y = lim x→+∞ x→+∞ x→+∞ 100 10000 1 48. y= x+1 x+2 x 1 , ln lim ln y = lim x→+∞ x→+∞ −1 lim y = e x+1 −x2 x + 2 = lim = −1; x→+∞ (x + 1)(x + 2) 1/x is the horizontal asymptote x→+∞ 0 50 0 (b) +∞ 49. (a) 0 (c) 50. (a) y = x 1+ln x ; lim+ ln y = lim+ (b) (c) y = (x + 1) (d) −∞ x→0 ln a x x→0 (f ) −∞ (ln a) ln x (ln a)/x = lim+ = lim+ ln a = ln a, lim+ y = eln a = a x→0 x→0 x→0 1 + ln x 1/x , lim ln y = lim x→0 (e) +∞ same as part (a) with x → +∞ ln a 0 x→0 (ln a) ln(x + 1) ln a = lim = ln a, lim y = eln a = a x→0 x + 1 x→0 x 51. 1 + 2 cos 2x x + sin 2x sin 2x does not exist, nor is it ±∞; lim = lim 1 + x→+∞ x→+∞ x→+∞ 1 x x 52. 2 2 − cos x 2x − sin x 2 − (sin x)/x does not exist, nor is it ±∞; lim = lim = x→+∞ 3 + cos x x→+∞ 3x + sin x x→+∞ 3 + (sin x)/x 3 53. 54. 55. 56. lim =1 lim x(2 + sin 2x) 2 + sin 2x = lim , which x→+∞ x→+∞ x→+∞ 1 + 1/x x+1 does not exist because sin 2x oscillates between −1 and 1 as x → +∞ lim (2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim sin x 11 + cos x + does not exist, nor is it ±∞; x2 2x x(2 + sin x) 2 + sin x = lim =0 lim x→+∞ x→+∞ x + 1/x x2 + 1 lim x→+∞ lim+ R→0 (a) (b) V t −Rt/L Le 1 = Vt L lim (π/2 − x) tan x = lim x→π/2 lim x→π/2 x→π/2 π/2 − x −1 = lim = lim sin2 x = 1 cot x x→π/2 − csc2 x x→π/2 1 − tan x = lim π/2 − x x→π/2 sin x 1 − π/2 − x cos x cos x − (π/2 − x) sin x (π/2 − x) cos x x→π/2 = lim −(π/2 − x) cos x −(π/2 − x) sin x − cos x x→π/2 = lim = lim x→π/2 (π/2 − x) sin x + cos x =0 −(π/2 − x) cos x + 2 sin x Supplementary Exercises 4 (c) 134 1/(π/2 − 1.57) ≈ 1255.765849, tan 1.57 ≈ 1255.765592; 1/(π/2 − 1.57) − tan 1.57 ≈ 0.000265 (c) kt − 1 (ln k )k t = lim = ln k + x→+∞ t→0 t→0 t 1 √ ln 0.3 = −1.20397, 1024 1024 0.3 − 1 = −1.20327; ln 2 = 0.69315, 1024 (a) No; sin(1/x) oscillates as x → 0. 57. (b) 58. lim x(k 1/x − 1) = lim + (b) √ 1024 2 − 1 = 0.69338 0.05 -0.35 0.35 -0.05 (c) For the limit as x → 0+ use the Squeezing Theorem together with the inequalities −x2 ≤ x2 sin(1/x) ≤ x2 . For x → 0− do the same; thus lim f (x) = 0. x→0 k + cos x = ±∞. Hence k = −1, and by the rule x2 2 √ −1 + cos x − sin...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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