# 49 twice once on 6 0 and once on 0 2 14 3317 15 let

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Unformatted text preview: lim = x→2 x − 2 t→0 t→0 t→0 t 2 4t 4 4 41. t = x − 1; sin(πx) = sin(πt + π ) = − sin πt; and lim 42. t = x − π/4; tan x − 1 = 43. −x ≤ x cos 45. x→π x→1 50π x sin(πx) sin πt = − lim = −π t→0 x−1 t tan x − 1 2 sin t 2 sin t ; lim = lim =2 t→0 t(cos t − sin t) cos t − sin t x→π/4 x − π/4 ≤x 44. −x2 ≤ x2 sin lim f (x) = 1 by the Squeezing Theorem x→0 46. 50π √ 3 x ≤ x2 lim f (x) = 0 by the Squeezing Theorem x→+∞ y 1 y = cos x -1 x 0 1 y = f (x) 4 y = 1 – x2 x -1 -1 47. 48. Let g (x) = − 1 1 sin x and h(x) = ; thus lim = 0 by the Squeezing Theorem. x→+∞ x x x y y x x 61 49. 50. Chapter 2 (a) πx sin x = sin t where x is measured in degrees, t is measured in radians and t = . Thus 180 sin x sin t π lim = lim = . x→0 x t→0 (180t/π ) 180 πx . Thus cos x = cos t where x is measured in degrees, t in radians, and t = 180 1 − cos x 1 − cos t = lim = 0. lim x→0 t→0 (180t/π ) x π π ≈ = 0.17453 18 18 51. (a) sin 10◦ = 0.17365 (b) sin 10◦ = sin 52. (a) cos θ = cos 2α = 1 − 2 sin2 (θ/2) ≈ 1 − 2(θ/2)2 = 1 − 1 θ2 2 (b) cos 10◦ = 0.98481 (c) cos 10◦ = 1 − 53. (a) 0.08749 (b) tan 5◦ ≈ 54. (a) h = 52.55 ft (b) πα is a good Since α is small, tan α◦ ≈ 180 approximation. (c) h ≈ 52.36 ft (a) Let f (x) = x − cos x; f (0) = −1, f (π/2) = π/2. By the IVT there must be a solution of f (x) = 0. 55. (b) 1 2 π 18 2 ≈ 0.98477 π = 0.08727 36 (c) 0.739 y 1.5 y=x 1 0.5 y = cos x 56. (a) x π/2 0 f (x) = x + sin x − 1; f (0) = −1, f (π/6) = π/6 − 1/2 > 0. By the IVT there must be a solution of f (x) = 0. (b) (c) x = 0.511. y y = 1– sin x y=x 0.5 0 57. π/6 x There is symmetry about the equatorial plane. (b) 58. (a) Let g (φ) be the given function. Then g (38) < 9.8 and g (39) > 9.8, so by the Intermediate Value Theorem there is a value c between 38 and 39 for which g (c) = 9.8 exactly. (a) does not exist (b) the limit is zero Supplementary Exercises 2 (c) 62 For part (a) consider the fact that given any δ > 0 there are inﬁnitely many rational numbers x satisfying |x| < δ and there are inﬁnitely many irrational numbers satisfying the same condition. Thus if the limit were to exist, it could not be zero because of the rational numbers, and it could not be 1 because of the irrational numbers, and it could not be anything else because of all the numbers. Hence the limit cannot exist. For part (b) use the Squeezing Theorem with +x and −x as the ‘squeezers’. CHAPTER 2 SUPPLEMENTARY EXERCISES 1. 1 (b) no limit (c) (f ) 2. (a) no limit 0 (g) 0 (h) 2 (a) f (x) = 2x/(x − 1) (d) 1 (i) (e) 3 1/2 y (b) 10 x 10 4. f (x) = −1 for a ≤ x < 5. (a) a+b a+b and f (x) = 1 for ≤ x ≤ b. 2 2 0.222 . . . , 0.24390, 0.24938, 0.24994, 0.24999, 0.25000; for x = 2, f (x) = the limit is 4. (b) 1.15782, 4.22793, 4.00213, 4.00002, 4.00000, 4.00000; to prove, use 6. (a) y=0 7. (a) x f(x) (b) 1 1.000 (b) 0.1 0.443 0.01 0.409 0.001 0.406 sin 4x 4 sin 4x tan 4x = = . x x cos 4x cos 4x 4x none 0.0001 0.406 1 , so the limit is 1/4; x+2 (c) 0.00001 0.405 y=2 0.000001 0.405 y 0.5 -1 8. (a) 1 x 0.4 amperes (b) (e) (d) 0.0187 9. y (a) [0.3947, 0.4054] 0.4 0.2 0.8 x 3 3 , 7.5 + δ 7.5 − δ It becomes inﬁnite. (b) 1 (c) Let g (x) = x − f (x). Then g (1) ≥ 0 and g (0) ≤ 0; by the Intermediate Value Theorem there is a solution c in [0, 1] of g (c) = 0. 63 10. Chapter 2 (a) (b) (c) (d) 1 − cos θ θ 1 − cos θ 1 − cos2 θ = tan lim = tan 0 = 0 θ→0 θ→0 θ→0 θ (1 + cos θ ) θ √ √ √ t−1 t−1 t−1 t+1 (t − 1)( t + 1) √ √ √ √ = t + 1; lim √ = = = lim( t + 1) = 2 t→1 t−1 t−1 t−1 t+1 t − 1 t→1 lim tan (3x2 = tan lim (2 − 1/x)5 (2x − 1)5 = → 25 /3 = 32/3 as x → +∞ 3 − 9x) + 2x − 7)(x (3 + 2/x − 7/x2 )(1 − 9/x2 ) sin(θ + π ) = sin θ cos π − cos θ sin π = − sin θ, so lim cos = cos lim θ→0 − sin θ 2θ θ →0 sin(θ + π ) 2θ = lim cos θ →0 − sin θ 2θ = cos − 1 2 11. If, on the contrary, f (x0 ) < 0 for some x0 in [0, 1], then by the Intermediate Value Theorem we would have a solution of f (x) = 0 in [0, x0 ], contrary to the hypothesis. 12. For x < 2 f is a polynomial and is continuous; for x > 2 f is a polynomial and is continuous. At x = 2, f (2) = −13 = 13 = lim+ f (x) so f is not continuous there. x→2 13. f (−6) = 185, f (0) = −1, f (2) = 65; apply Theorem 2.4.9 twice, once on [−6, 0] and once on [0, 2] 14. 3.317 15. Let = f (x0 )/2 > 0; then there corresponds δ > 0 such that if |x − x0 | < δ then |f (x) − f (x0 )| < , − < f (x) − f (x0 ) < , f (x) > f (x0 ) − = f (x0 )/2 > 0 for x0 − δ < x < x0 + δ . 16. y 1 4 x 1.449 (x must be ≥ −1) (b) x = 0, ±1.896 17. (a) 18. Since lim sin(1/x) does not exist, no conclusions can be drawn. 19. (a) 5, no limit, no limit 20. (a) −1/5, +∞, −1/10, −1/10, no limit, 0, 0 21. a/b 24. 2 x→0 √...
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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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