# 1 cos 5h1 cos 5h1 cos 7h 1 cos 5h cos 7h 1 cos

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Unformatted text preview: 58 Since lim p(x) = −∞ and lim p(x) = +∞ (or vice versa, if the leading coeﬃcient of p is negative), x→−∞ x→+∞ it follows that for M = −1 there corresponds N1 < 0, and for M = 1 there is N2 > 0, such that p(x) < −1 for x < N1 and p(x) > 1 for x > N2 . Choose x1 < N1 and x2 > N2 and use Theorem 2.4.9 on the interval [x1 , x2 ] to ﬁnd a solution of p(x) = 0. 43. For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f (−1.3) < 0 and f (−1.2) > 0, the midpoint x = −1.25 of [−1.3, −1.2] is the required approximation of the root. For the positive root use the interval [0, 1]; since f (0.7) < 0 and f (0.8) > 0, the midpoint x = 0.75 of [0.7, 0.8] is the required approximation. 44. x = −1.25 and x = 0.75. 10 1 0.7 -2 0.8 -1 -5 -1 45. For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f (−1.7) < 0 and f (−1.6) > 0, use the interval [−1.7, −1.6]. Since f (−1.61) < 0 and f (−1.60) > 0 the midpoint x = −1.605 of [−1.61, −1.60] is the required approximation of the root. For the positive root use the interval [1, 2]; since f (1.3) > 0 and f (1.4) < 0, use the interval [1.3, 1.4]. Since f (1.37) > 0 and f (1.38) < 0, the midpoint x = 1.375 of [1.37, 1.38] is the required approximation. 46. x = −1.605 and x = 1.375. 1 -1.7 1 -1.6 1.3 -2 47. 48. 1.4 -0.5 x = 2.24. b a + . Since lim+ f (x) = +∞ and lim− f (x) = −∞ there exist x1 > 1 and x2 < 3 x→1 x→3 x−1 x−3 (with x2 > x1 ) such that f (x) > 1 for 1 < x < x1 and f (x) < −1 for x2 < x < 3. Choose x3 in (1, x1 ) and x4 in (x2 , 3) and apply Theorem 2.4.9 on [x3 , x4 ]. Set f (x) = 49. The uncoated sphere has volume 4π (x − 1)3 /3 and the coated sphere has volume 4πx3 /3. If the volume of the uncoated sphere and of the coating itself are the same, then the coated sphere has twice the volume of the uncoated sphere. Thus 2(4π (x − 1)3 /3) = 4πx3 /3, or x3 − 6x2 + 6x − 2 = 0, with the solution x = 4.847 cm. 50. Let g (t) denote the altitude of the monk at time t measured in hours from noon of day one, and let f (t) denote the altitude of the monk at time t measured in hours from noon of day two. Then g (0) < f (0) and g (12) > f (12). Use Exercise 36. 51. We must show lim f (x) = f (c). Let x→c > 0; then there exists δ > 0 such that if |x − c| < δ then |f (x) − f (c)| < . But this certainly satisﬁes Deﬁnition 2.3.3. 59 Chapter 2 EXERCISE SET 2.5 3. x = nπ , n = 0, ±1, ±2, . . . 2. x = π 1. none 4. x = nπ + π/2, n = 0, ±1, ±2, . . . 6. none 9. 2nπ + π/6, 2nπ + 5π/6, n = 0, ±1, ±2, . . . 11. 5. x = nπ , n = 0, ±1, ±2, . . . (a) 13. cos 10. (c) x3 , cos x, x + 1 (e) (f ) x5 − 2x3 + 1, cos x sin x, sin x Use Theorem 2.4.6. lim x→+∞ 1 x (b) = cos 0 = 1 πx 15. sin lim x→+∞ 2 − 3x π = sin − 3 none (b) |x|, sin x sin x, x3 + 7x + 1 √ x, 3 + x, sin x, 2x (d) (a) 12. 8. x = nπ + π/2, n = 0, ±1, ±2, . . . 7. none g (x) = cos x, g (x) = sin 16. √ 3 =− 2 lim 2 x 14. x2 1 , g (x) = x2 + 1 +1 sin h 1 1 lim = 2 h→0 h 2 x→+∞ 1 θ = sin 0 = 0 sin θ = +∞ θ sin 3θ =3 θ→0 3θ 18. 19. sin x − lim− = −1 x→0 x 20. 1 3 21. √ sin x 1 lim+ x lim+ =0 x→0 5 x→0 x 22. sin 6x 6 sin 6x 8x 6 3 sin 6x = , so lim == x→0 sin 8x sin 8x 8 6x sin 8x 8 4 23. tan 7x 7 sin 7x 3x 7 7 tan 7x = so lim = (1)(1) = x→0 sin 3x sin 3x 3 cos 7x 7x sin 3x 3(1) 3 17. 3 lim sin θ =0 θ→0 θ lim sin θ lim 24. θ→0 lim θ →0 + lim θ →0 + sin x x→0 x 2 lim = 1 3 h =1 h→0 sin h lim cos h lim 25. h→0 26. sin h 1 + cos h sin h(1 + cos h) 1 + cos h sin h = = = ; no limit 1 − cos h 1 − cos h 1 + cos h 1 − cos2 h sin h 27. θ2 (1 + cos θ) θ2 1 + cos θ = = 1 − cos θ 1 + cos θ 1 − cos2 θ 28. cos( 1 π − x) = sin( 1 π ) sin x = sin x, so lim 2 2 29. 0 31. θ sin θ x→0 2 θ2 = (1)2 2 = 2 θ→0 1 − cos θ (1 + cos θ) so lim x cos 1 2π −x 30. (1 − cos 5h)(1 + cos 5h)(1 + cos 7h) 1 − cos 5h = cos 7h − 1 (cos 7h − 1)(1 + cos 5h)(1 + cos 7h) 1 − cos 5h 25 =− lim h→0 cos 7h − 1 49 =1 t2 = 1 − cos2 t =− 25 49 t sin t 2 sin 5h 5h 2 , so lim t→0 7h sin 7h t2 =1 1 − cos2 t 2 1 + cos 7h so 1 + cos 5h Exercise Set 2.5 32. 34. 36. 37. 60 1 = lim sin t; t→+∞ x limit does not exist lim sin 33. x→0+ lim x − 3 lim x→0 x→0 sin x = −3 x k = f (0) = lim x→0 2 + lim x→0 sin x =3 x sin 3x sin 3x = 3 lim = 3, so k = 3 x→0 3x x lim f (x) = k lim x→0 x→0− 35. 1 = lim cos t; t→+∞ x limit does not exist lim cos x→0+ 1 sin kx = k , lim+ f (x) = 2k 2 , so k = 2k 2 , k = x→0 kx cos kx 2 38. No; sin x/|x| has unequal one-sided limits. 39. (a) lim t→0+ sin t =1 t (b) lim t→0− 1 − cos t = 0 (Theorem 2.5.3) t π−x t = lim =1 t→0 sin t sin x (c) sin(π − t) = sin t, so lim 40. cos π π cos(π/x) (π − 2t) sin t π − 2t sin t − t = sin t, so lim = lim = lim...
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