Chpter 1 - C H A P T E R 1 Limits and Their Properties...

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Unformatted text preview: C H A P T E R 1 Limits and Their Properties Section 1.1 Section 1.2 Section 1.3 Section 1.4 Section 1.5 A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 46 Finding Limits Graphically and Numerically . . . . . . . . 47 Evaluating Limits Analytically . . . . . . . . . . . . . . . 57 Continuity and One-Sided Limits . . . . . . . . . . . . . . 68 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Review Exercises Problem Solving C H A P T E R 1 Limits and Their Properties Section 1.1 A Preview of Calculus 300 feet 2. Calculus: velocity is not constant Distance 3. Calculus required: slope of tangent line at x change, and equals about 0.16. 1 2 bh 1 2 15 2 1. Precalculus: 20 ft sec 15 seconds 20 ft sec 15 seconds slope 300 feet 0.08 2 is rate of 4. Precalculus: rate of change 5. Precalculus: Area 5 3 sq. units 6. Calculus required: Area bh 2 2.5 5 sq. units 7. f x (a) 4 3 y 4x x2 8. f x (a) 2 y x P(4, 2) P x x 1 2 3 1 2 3 4 5 (b) slope m 4x x x x2 1 1 3 x 1 2 1 1.5 0.5 3 x (b) slope 3 x, x 1 x m x x x 2 4 x 2 1 3 0.2679 0.2361 1 4 2 1 4 0.25. 2 x 1 2 x 2 , x 4 x x x 2: m 1.5: m 0.5: m 3 3 3 1: m 3: m 5: m 1.5 2.5 x x 1 1 2 1 3 2 1 5 2 (c) At P 1, 3 the slope is 2. You can improve your approximation of the slope at x 1 by considering x-values very close to 1. (c) At P 4, 2 the slope is You can improve your approximation of the slope at x 4 by considering x-values very close to 4. 5 2 5 3 5 1.5 5 4 5 2 9. (a) Area Area 5 1 5 2 10.417 5 2.5 5 3 5 3.5 5 4 5 4.5 9.145 (b) You could improve the approximation by using more rectangles. 46 Section 1.2 Finding Limits Graphically and Numerically 47 10. (a) For the figure on the left, each rectangle has width Area sin 2 2 2 1 4 sin 1 sin 3 4 sin 4 . 4 4 4 2 2 2 1.8961 . sin For the figure on the right, each rectangle has width Area sin sin 3 2 sin 3 2 sin 1 2 2 3 5 6 6 6 6 3 1 2 1 6 2 3 2 6 1.9541 (b) You could obtain a more accurate approximation by using more rectangles. You will learn later that the exact area is 2. 11. (a) D1 (b) D2 5 1 2.693 1 2 1 1 5 2 16 5 2 5 2 3 16 1 6.11 5.66 5 3 5 2 4 5 2 2 1 5 4 1 2 1.302 1.083 1.031 (c) Increase the number of line segments. Section 1.2 1. x f x lim 1.9 0.3448 x x2 x 2 Finding Limits Graphically and Numerically 1.99 0.3344 1.999 0.3334 2.001 0.3332 2.01 0.3322 2.1 0.3226 x2 2 0.3333 Actual limit is 1 . 3 2. x f x lim 1.9 0.2564 x x2 2 4 1.99 0.2506 0.25 1.999 0.2501 2.001 0.2499 2.01 0.2494 2.1 0.2439 x2 Actual limit is 1 . 4 3. x f x lim 0.1 0.2911 x 3 x 0.01 0.2889 3 0.2887 0.001 0.2887 0.001 0.2887 0.01 0.2884 0.1 0.2863 x0 Actual limit is 1 2 3 . 48 4. Chapter 1 Limits and Their Properties x f x lim 3.1 0.2485 1 x x 2 3 3.01 0.2498 0.25 3.001 0.2500 2.999 0.2500 1 4. 2.99 0.2502 2.9 0.2516 x 3 Actual limit is 5. x f x lim 2.9 0.0641 1 x x 1 3 2.99 0.0627 1 4 2.999 0.0625 0.0625 3.001 0.0625 Actual limit is 3.01 0.0623 1 16 . 3.1 0.0610 x3 6. x f x lim 3.9 0.0408 x x x 1 3.99 0.0401 4 5 4 3.999 0.0400 0.04 4.001 0.0400 4.01 0.0399 1 4.1 0.0392 x4 Actual limit is 25 . 7. x f x lim 0.1 0.9983 sin x x 1.0000 0.01 0.99998 0.001 1.0000 0.001 1.0000 0.01 0.99998 0.1 0.9983 x0 (Actual limit is 1.) (Make sure you use radian mode.) 8. x f x lim 0.1 0.0500 cos x x x 1 lim 4 x 1 0.01 0.0050 0.0000 0.001 0.0005 0.001 0.0005 0.01 0.0050 0.1 0.0500 x0 (Actual limit is 0.) (Make sure you use radian mode.) 10. lim x2 x1 9. lim 4 x3 2 3 2 3 11. lim f x x2 x2 2 12. lim f x x1 x1 lim x2 13. lim x 5 does not exist. For values of x to the left of 5, x5 x 5 x 5 x 5 equals 1, whereas for values of x to the right of 5, x 5 x 5 equals 1. 14. lim 1 does not exist since the function increases and x 3 decreases without bound as x approaches 3. x3 15. lim sin x x1 0 16. lim sec x x0 1 17. lim cos 1 x does not exist since the function oscillates x0 18. lim tan x does not exist since the function increases and x 2 between 1 and 1 as x approaches 0. decreases without bound as x approaches 2. Section 1.2 19. (a) f 1 exists. The black dot at 1, 2 indicates that f 1 2. (b) lim f x does not exist. As x approaches 1 from the x1 Finding Limits Graphically and Numerically (c) f 4 does not exist. The hollow circle at 4, 2 indicates that f is not defined at 4. 49 (d) lim f x exists. As x approaches 4, f x approaches 2: x4 x4 left, f x approaches 2.5, whereas as x approaches 1 from the right, f x approaches 1. 20. (a) f 2 does not exist. The vertical dotted line indicates that f is not defined at 2. (b) lim f x does not exist. As x approaches x 2 lim f x 2. (e) f 2 does not exist. The hollow circle at 2, 1 2 indicates that f 2 is not defined. 1 (f) lim f x exists. As x approaches 2, f x approaches 2 : x2 x2 2, the values of f x do not approach a specific number. (c) f 0 exists. The black dot at 0, 4 indicates that f 0 4. (d) lim f x does not exist. As x approaches 0 from the left, f x approaches 1, whereas as x approaches 0 2 from the right, f x approaches 4. x0 lim f x 1 2. (g) f 4 exists. The black dot at 4, 2 indicates that f 4 2. (h) lim f x does not exist. As x approaches 4, the values x4 of f x do not approach a specific number. 21. lim f x exists for all c xc 3. In particular, lim f x x2 2. 22. lim f x exists for all c xc 2, 0. In particular, lim f x x 4 2. 23. 6 5 4 3 2 1 -2 -1 -1 -2 y 24. 2 y 1 f - 2 -1 x 1 2 3 4 5 2 x xc lim f x exists for all values of c y 6 5 4 4. xc lim f x exists for all values of c y 4 3 . 25. One possible answer is 26. One possible answer is f 2 1 -3 -2 -1 -1 x 1 2 2 1 -2 -1 -1 x 1 2 3 4 5 27. C t (a) 3 0.75 0.50 t 1 (b) t 3 3.3 2.25 2.25 2.5 1.75 3.4 2.25 3.5 2.25 3.6 2.25 3.7 2.25 4 2.25 C 1.75 t3.5 0 0 5 lim C t t C 2 1.25 (c) 2.9 1.75 3 1.75 3.1 2.25 3.5 2.25 4 2.25 3. lim C t does not exist. The values of C jump from 1.75 to 2.25 at t t3 50 Chapter 1 0.35 1 Limits and Their Properties 0.12 t 1 28. C t (a) 0 0 5 (b) t Ct lim C t t Ct 3 0.59 3.3 0.71 3.4 0.71 3.5 0.71 3.6 0.71 3.7 0.71 4 0.71 t3.5 0.71 2 0.47 2.5 0.59 2.9 0.59 3 0.59 3.1 0.71 3.5 0.71 4 0.71 3. 0.4. If 0 < x 2 < 0.4, then (c) lim C t does not exist. The values of C jump from 0.59 to 0.71 at t t3 29. We need f x x 2 x 3 1 x 3 1 x 1 f x 3 x 2 < 0.4. Hence, take 3 < 0.4, as desired. 2 x x < 0.01. Let 1 1 < 1 102 101 1 101 30. We need f x 1 < x 101 1 2 < 1 1 1 . If 0 < x 101 2 < 1 , then 101 1 1 101 1 < x 101 1 < 100 101 100 < x 101 1 > x and we have f x 1 1 x 1 1 2 x x 1 101 < 1 100 101 1 100 0.01. 31. You need to find 1 f x 1 x 0.1 < 1 0.1 < 9 < 10 10 > 9 10 9 1 > x 1 > x 9 1 x such that 0 < x 1 < 0.1. That is, 1 < 0.1 1 x 1 x x < 1 < > 1 < implies So take 1 . Then 0 < x 11 1 < 1 11 1 < implies 1 < x 11 1 < x 11 1 1 < . 9 0.1 Using the first series of equivalent inequalities, you obtain f x 1 1 x 1 < 0.1. 11 10 10 11 10 11 1 . 11 1 1 > 1 > Section 1.2 32. You need to find such that 0 < x 2 < implies f x 3 x2 1 3 x2 4 < 0.2. That is, 0.2 0.2 3.8 3.8 3.8 2 4 So take Then 0 < x 4.2 3.8 < 4 < x2 < x2 < x < x 2 Finding Limits Graphically and Numerically 2 2 3x 3x 0 < x 8 L 51 33. lim 3x x2 3x 8 < 0.01 6 < 0.01 2 < 0.01 2 < 0.01 3 2 < 2 < 0.01 6 < 0.01 8 < 0.01 L < 0.01 0.0033 0.01 , you have 3 x2 < 0.2 < 4 0.2 < 4.2 < 4.2 < 4.2 2 4.2 2 < 2 0.0494. implies 2 < 2 < 4.2 4.2 2 2. Hence, if 0 < x 3x 3x 3x 2 f x 2 < x 2 < x Using the first series of equivalent inequalities, you obtain f x x 2 x 2 2 1 x 2 0 < x 3 x2 4 < 0.2. 34. lim 4 x4 2 L 35. lim x2 x2 3 3 x2 1 L x2 2 < 0.01 x < 0.01 2 4 < 0.01 1 < 0.01 4 < 0.01 2 < 0.01 4 x x 2 x 2 x x 2 < 0.01 2 < 0.01 x 2 0.01 5 0.002. 4 < 0.02 4 < < 0.01 Hence, if 0 < x 1 x 2 2 4 x 2 f x 4 0.02, you have If we assume 1 < x < 3, then Hence, if 0 < x x 2 < 2 < 0.002 2 < 0.01 4 < 0.01 1 < 0.01 L < 0.01 0.002, you have 1 1 0.01 < 0.01 5 x 2 x < 0.01 2 2 < 0.01 x 2 x x2 x2 L < 0.01 3 f x 36. lim x2 x5 4 4 x 2 29 L Hence, if 0 < x x x 5 x x2 x2 0.01 11 0.0009. 4 f x 5 < 5 < 0.01 , you have 11 x2 29 < 0.01 25 < 0.01 5 < 0.01 0.01 1 < 0.01 11 x 5 x 5 x x 5 < 0.01 25 < 0.01 29 < 0.01 L < 0.01 5 < 0.01 x 5 If we assume 4 < x < 6, then 52 Chapter 1 3 > 0: 3 x Hence, let Hence, if 0 < x x x 3 f x 5 Limits and Their Properties 38. lim 2x x 3 37. lim x x2 5 > 0: 5 2x 2x 1 Given x Given 5 < 2 < . 2 < , you have 2x 1 < 6 < 3 < 3 < 2. 3 < 3 < 6 < 1 < 2 < 5 < L < . Hence, let x 2 Hence, if 0 < x x 2x 2x 5 f x 2 2 , you have L < . 39. lim x 4 1 2x 1 > 0: 1 1 2x 1 2 1 2 4 1 3 40. lim x1 2 3x 9 > 0: 9 2 3x 2 3 2 3 1 9 29 3 Given 1 2x Given 3 < 2 3x 29 3 2 3 < < 2 < 4 4 < < 2 x x x 3 2 1 < x Hence, let 1 < Hence, let 2 . 4 < 2 < 3 2 . 1 < 1 < 2 3 29 3 3 2 3 2 Hence, if 0 < x x 1 2x 1 2x 2 , you have Hence, if 0 < x x 2 3x 2 3x , you have 4 2 < 3 < L < . 42. lim 1 < < 1 f x 9 f x L < . 41. lim 3 x6 3 > 0: 3 < 0 < x2 1 > 0: 1 1 < Given 3 Given 0 < Hence, any > 0 will work. > 0, you have Hence, any > 0 will work. > 0, you have Hence, for any 1 f x Hence, for any 3 f x 3 < 1 < L < . L < . Section 1.2 43. lim 3 x x0 Finding Limits Graphically and Numerically x 4 > 0: x 2 2 x x x 2 < 2 < 4 < x x 2 2 53 0 > 0: 3 44. lim x 3 x4 Given 0 < x < x < 3 Given Hence, let Hence for 0 x 3. Assuming 1 < x < 9, you can choose 3, 3 3 . Then, x 2 0 you have 0 < x 4 < 3 x x 4 < 2 < . x < 3 3 x < 0 < L < . x f x 45. lim x x 2 2 > 0: 2 2 4 46. lim x x3 3 > 0: 0 Given Given x x 2 2 x 4 < 4 < 2 < x x 2<0 Hence, let 3 x 0 < 3 < . 3 < , you have x Hence, let 2 . x 2 Hence for 0 < x 2 2 < < Hence for 0 < x x x x x 2 2 f x 47. lim x2 x1 , you have x x 3 f x 3 < 0 < L < . 2 < 4 < 4 < L < . 2 48. lim x2 x 3 (because x 2 < 0) 1 > 0: 3x > 0: 0 Given Given 2 < 1 < 1 < x2 1 x2 x2 3x xx x 0 < 3 < x 1 x x 3 < x 2, then < 1 < x 1 3. 3 , you have If we assume 4 < x < 3 4. 4 , you have If we assume 0 < x < 2, then Hence for 0 < x x x2 x2 1 f x 1 < Hence for 0 < x x xx x2 3x f x 3 < 3 < 1 1 < 4 x 1 < 1 < 2 < 1 1 < 3 x 1 0 < L < . 2 < . 54 Chapter 1 x Limits and Their Properties 0.5 49. f x lim f x 5 3 x 4 1 6 -6 50. f x 6 -0.1667 x x2 1 2 4x 3 3 -3 4 5 x4 x3 lim f x -4 The domain is 5, 4 4, not show the hole at 4, 1 . 6 x x 6 0 0 . The graphing utility does The domain is all x 1, 3. The graphing utility does not show the hole at 3, 1 . 2 x x2 3 9 1 6 -9 51. f x lim f x 9 3 10 52. f x lim f x 3 x9 3 x3 10 -3 The domain is all x 0 except x 9. The graphing utility does not show the hole at 9, 6 . The domain is all x 3. The graphing utility does not 1 show the hole at 3, 6 . 53. lim f x x8 25 means that the values of f approach 25 as x gets closer and closer to 8. 56. (i) The values of f approach different numbers as x approaches c from different sides of c: y 4 3 2 1 -4 -3 -2 -1 -1 -3 -4 x 1 2 3 4 54. No. The fact that f 2 4 has no bearing on the existence of the limit of f x as x approaches 2. 55. No. The fact that lim f x x2 4 has no bearing on the value of f at 2. (ii) The values of f increase without bound as x approaches c: y 6 5 4 3 2 1 -3 -2 -1 -1 -2 x 2 3 4 5 (iii) The values of f oscillate between two fixed numbers as x approaches c: y 4 3 -4 -3 -2 x 2 3 4 -3 -4 57. (a) C r (b) If C If C 2 r C 2 6 2 5.5, r 6.5, r 3 5.5 2 6.5 2 0.9549 cm 0.87535 cm 1.03451 cm 58. V 4 3 r ,V 3 2.48 (a) 2.48 r3 r (b) 4 3 r 3 1.86 0.8397 in. V 2.51 Thus 0.87535 < r < 1.03451 (c) lim 2 r r 3 2.45 2.45 0.5849 0.8363 6; 0.5; 0.0796 4 3 r 2.51 3 r 3 0.5992 r 0.8431 (c) For 2.51 2.48 0.03, 0.003 Section 1.2 59. f x x0 1 x Finding Limits Graphically and Numerically 55 1 x y 7 x 1 x x e 2.71828 0.1 0.01 0.001 0.0001 0.00001 0.000001 x f x 2.867972 2.731999 2.719642 2.718418 2.718295 2.718283 x 0.1 0.01 0.001 0.0001 0.00001 0.000001 f x 2.593742 2.704814 2.716942 2.718146 2.718268 2.718280 lim 1 3 2 1 -3 -2 -1 -1 (0, 2.7183) 1 2 3 4 5 60. f x x f x lim f x x 1 x x 1 3 y 1 2 2 0.5 2 0.1 2 0 Undef. 0.1 2 0.5 2 1.0 2 -2 -1 -1 1 x 1 2 x0 Note that for 61. 1 < x < 1, x 0, f x x 1 x x 1 2. 62. 0.002 0.005 (1.999, 0.001) (2.001, 0.001) 1.998 0 2.002 2.99 0 3.01 Using the zoom and trace feature, 0 < x 2 < 0.001, x2 x 4 2 0.001. That is, for 4 < 0.001. From the graph, 3 ,3 Note: x2 x 3x 3 0.001. Thus 2.999, 3.001 . x for x 3. 63. False; f x sin x x is undefined when x From Exercise 7, we have lim sin x x 1. 0. 64. True x0 65. False; let f x f 4 x4 66. False; let x 4x, 10, 2 x x 4 . 4 x4 f x lim f x x2 10, x4 4x, x x 4x 4 4 . 10 0 10 x4 lim x2 0 and f 4 lim f x x x0.25 lim x2 4x 0 10 67. f x (a) lim x 0.5 is true. 1 4, (b) lim f x x approaches x0 x 0 is false. As x approaches 0.25 1 2 0.5. x is not defined on an open interval containing 0 f x because the domain of f is x 0. 56 Chapter 1 Limits and Their Properties 68. The value of f at c has no bearing on the limit as x approaches c. 69. If lim f x xc L1 and lim f x xc 1 L2, then for every and x L2 c < L1 2 > 0, there exists 1 > 0 and 2 > 0 such that 1 x c < f x L1 < f x f x L2 < . Let L2 L1 L2. equal the smaller of f x f x L2 < and . 2. Then for x Therefore, L1 c < , we have L1 L2 < 2 . Since f x > 0 is arbitrary, it follows that L1 70. f x mx b, m c < 0. Let > 0 be given. Take m . L 0 means that for every 71. lim f x xc exists > 0 such that if 0 < x then f x L 0 < . L < c < , > 0 there If 0 < x m m x mx , then c < mc < b < mx b mc xc This means the same as f x mc b. 0 < x Thus, lim f x xc when which shows that lim mx b c < . L. 72. (a) 3x 1 3x 1 x2 0.01 9x2 9x4 1 x2 x2 1 100 1 100 1 90x2 1 (b) We are given lim g x xc L > 0. Let c < 1 2 L. There exists gx > 0 such that 0 < x implies that L < L < gx 2 L < 2 gx L . That is, 2 L < < 1 10x2 100 Thus, 3x 1 3x 1 x2 0.01 > 0 if 10x2 1 < 0 and 90x2 1 < 0. Let a, b 1 , 90 1 . 90 L 2 3L 2 ,x c, we have For all x 0 in a, b , the graph is positive. You can verify this with a graphing utility. 73. Answers will vary. x2 x ,c For x in the interval c L g x > > 0, as desired. 2 74. lim x 4 12 x4 7 n 1 2 3 4 4 4.1 0.1 n f 4 7.1 0.1 n n 1 2 3 4 4 3.9 0.1 n f 4 6.9 6.99 0.1 n 4.01 4.001 4.0001 7.01 7.001 7.0001 3.99 3.999 3.9999 6.999 6.9999 Section 1.3 75. The radius OP has a length equal to the altitude z of the h h triangle plus . Thus, z 1 . 2 2 Area triangle Area rectangle 1 b 1 2 bh h 2 1 h 2 5 h 2 h bh 2h 1 2 5 h 2 h P Evaluating Limits Analytically 57 O b 1 Since these are equal, b 1 2 76. Consider a cross section of the cone, where EF is a diagonal of the inscribed cube. AD 3, BC 2. Let x be the length of a side of the cube. Then EF x 2. By similar triangles, EF BC x 2 2 AG AD 3 3 3 2x 3 2 2x x 6 6 6 3 2 2 9 2 7 6 0.96. 2x x B E A G F D C Solving for x, Section 1.3 1. -8 7 Evaluating Limits Analytically (a) lim h x x5 13 0 6 2. 10 (a) lim g x x4 2.4 4 (b) lim h x x 1 (b) lim g x x0 0 10 -7 -5 hx x2 5x gx 12 x x 9 3 3. - 4 (a) lim f x x0 x 0 0.524 4. -5 10 (a) lim f t t4 0 5 (b) lim f x 3 (b) lim f t 10 t 1 -4 6 f t - 10 f x x cos x tt 4 58 Chapter 1 24 Limits and Their Properties 16 6. lim x3 x 2 5. lim x4 x2 2 3 8 7. lim 2x x0 1 3x 4x 20 3 1 1 2 1 3 3 2 8. lim 3x x 3 2 3 3 2 7 9. lim x2 x 3 9 4 3 9 1 0 18 12 1 10. lim 7 x1 x2 1 1 2 1 0 11. lim 2x2 x 3 2 3 12. lim 3x3 x1 2x2 4 31 3 21 2 4 5 13. lim x2 1 x x x2 1 2 3 4 5x x 2 1 12 3 4 57 7 2 2 5 35 9 2 5 35 3 14. lim 2 x 2 3 5 3 23 3 2 2 3 5 3 8 x 3 2 15. lim x1 16. lim x3 2x x 17. lim x7 18. lim x x x3 1 4 4 3 3 3 1 3 4 4 4 2 3 2 19. lim x3 x 1 3 1 2 20. lim 3 x x4 21. lim x x 4 3 2 4 3 2 1 22. lim 2x x0 1 20 1 1 23. (a) lim f x x1 5 43 1 64 g f 1 4 (b) lim g x x4 (c) lim g f x x1 g4 64 24. (a) lim f x x 3 x4 3 42 16 g4 7 4 25. (a) lim f x x1 4 3 1 1 3 2 2 (b) lim g x (c) lim g f x x 3 (b) lim g x x3 16 (c) lim g f x x1 g3 26. (a) lim f x x4 2 42 3 34 6 3 3 1 21 27. lim sin x x 2 sin 2 1 (b) lim g x x21 21 g 21 (c) lim g f x x4 28. lim tan x x tan 0 29. lim cos x2 x 3 cos 2 3 1 2 30. lim sin x1 x 2 sin 2 5 6 7 6 1 31. lim sec 2x x0 sec 0 1 32. lim cos 3x x cos 3 1 33. x5 lim sin x 6 sin 1 2 2 3 3 34. x5 lim cos x 3 cos 5 3 1 2 35. lim tan x3 x 4 tan 3 4 1 36. lim sec x7 x 6 sec Section 1.3 Evaluating Limits Analytically 3 2 59 37. (a) lim 5g x xc 5 lim g x xc 53 15 xc 38. (a) lim 4f x xc 4 lim f x xc 4 6 lim g x xc (b) lim f x xc gx xc lim f x lim g x 2 2 3 3 6 5 (b) lim f x xc gx xc lim f x 3 2 3 2 1 2 1 2 3 4 2 (c) lim f x g x xc xc lim f x lim g x xc (d) lim xc f x gx lim f x xc xc (c) lim f x g x xc lim g x 2 3 xc lim f x lim g x xc (d) lim f x xc g x xc xc lim f x lim g x 3 2 1 2 3 39. (a) lim f x xc 3 xc lim f x 3 4 3 64 2 12 4 3 2 40. (a) lim 3 f x xc 3 xc xc lim f x 27 18 2 3 27 3 2 27 2 3 (b) lim xc f x xc lim f x 4 34 3 2 (b) lim f x xc 18 2 lim f x xc lim 18 xc (c) lim 3 f x xc 3 lim f x xc 3 2 (c) lim f x xc lim f x 729 2 3 (d) lim f x xc xc lim f x 8 (d) lim f x xc 2 3 xc lim f x 2 3 27 9 41. f x x 0. x0 2x 1 and g x lim f x x 1 2x2 x 1 3 x agree except at 42. f x x 3 and h x lim f x x2 x 3x agree except at x 0. (a) lim g x (b) lim g x x 1 (a) lim h x x 2 x0 x 2 5 3 lim f x (b) lim h x x0 x0 lim f x 43. f x xx 1 and g x lim f x x 1 x3 x 2 0 x agree except at x 1 1. 44. g x 1 x x1 1 and f x x x2 x agree except at x 0. (a) lim g x x1 (a) lim f x does not exist. (b) lim f x x0 x1 (b) lim g x x 1 lim f x 1 45. f x x2 x x 1 1 and g x 1 x 1 x 1 agree except at x 2 1. 46. f x x 2x2 x 1. x 1 x 1 3 and g x 2x 3 agree except at lim f x 3 lim g x lim f x 4 x 1 lim g x 5 -3 4 -8 4 -4 -8 60 Chapter 1 x3 x 2. x2 Limits and Their Properties x3 x 1. x 1 47. f x x 8 and g x 2 lim g x x2 2x 4 agree except at 48. f x x 1 and g x 1 lim g x x2 x 1 agree except at lim f x 12 x2 12 lim f x 7 x 1 3 -9 0 9 -4 -1 4 49. lim x5 x x2 5 25 x5 lim x 1 x 5 5 x 5 1 10 5 50. lim x2 2 x2 x 4 x2 lim x x x 2 2 x 2 1 2 x x x x 1 4 4 x 4 x 1 2 3 6 1 2 1 2 x5 lim x x2 lim 51. lim x2 x2 x 9 6 x 3 x 3 lim x x x x 3 x 3 x 2 3 5 6 2 3 5 6 52. lim x4 x2 x2 5x 2x 4 8 x4 lim x 3 lim x4 lim 53. lim x x0 5 x 5 x0 lim x x x 5 x 5 5 5 5 5 x x 5 5 x0 5 5 x 1 5 1 5 2 5 5 10 lim x0 x lim 54. lim 2 x0 x x 2 x0 lim 2 2 2 x x x 4 x x x x x x 2 2 2 x x x 9 5 x x 3 2 2 x x x0 2 2 2 1 x 2 1 2 2 2 4 x0 lim lim 55. lim x x x4 5 3 4 x4 lim 5 3 4 5 x 1 2 3 5 5 3 3 x4 x4 lim x lim 1 x 5 3 x 1 9 3 3 x 1 2 1 6 1 1 1 4 56. lim x x 1 x3 1 2 3 1 3 x3 lim 1 1 2 2 x3 lim x 3 x3 lim x 2 57. lim 3 x0 x x 1 x0 lim 3 3 4 3 x x3x x 4 4 x 1 x0 lim 3 x x 3 x x0 lim 3 1 x3 1 9 58. lim x x0 4 x 1 4 x0 lim 4x 59. lim 1 16 2x x0 x x 2x lim 2x x0 2 x x 2 2x lim 2 x0 x0 lim 4x 4 Section 1.3 x x2 x x 2 Evaluating Limits Analytically 61 60. lim x2 x0 lim x2 2x x x 1 x2 2x x 2 x2 x0 lim x 2x x 2x x x x0 lim 2x x0 x 2x 61. lim x 2x x x 1 x0 lim x2 x 2 2x x 2 x 1 x2 2x 1 x0 lim 2x x0 x 2 2x 2 62. lim x x0 x3 x x3 lim lim x3 3x2 x x0 3x x x 2 x 2 3 x3 x 3x2 x0 3x x x x lim 3x2 x0 3x x x 2 3x2 63. lim x x0 2 x 0.1 0.358 2 0.354 -3 2 x f x 0.01 0.354 x 2 x 2 0.001 0.354 0 ? x x x 0.001 0.354 2 x 2 2 2 2 0.01 0.353 x x 2 0.1 0.349 2 2 x0 3 -2 Analytically, lim x0 x0 lim 2 2 x 1 2 1 2 2 2 2 4 0.354. x0 lim x lim 64. f x 4 x 15.9 x 16 0 1 20 x f x 15.99 0.125 x 16 15.999 0.125 lim 4 x 1 x 4 16 ? 16.001 0.125 x x 1 . 8 16.01 0.125 16.1 0.1248 -1 0.1252 4 x16 x It appears that the limit is 4 0.125. Analytically, lim x16 4 x16 lim 1 65. lim 2 x0 x x 1 2 3 1 4 -5 1 x f x 0.1 0.263 1 0.01 0.251 1 2 0.001 0.250 0 ? 0.001 0.250 0.01 0.249 0.1 0.238 -2 Analytically, lim 2 x0 x x x0 lim 2 22 2 x x 1 x x0 lim x 22 x 1 x x0 lim 1 22 x 1 . 4 62 Chapter 1 x5 x 32 2 1.9 72.39 Limits and Their Properties 100 66. lim x2 80 x f x 1.99 79.20 x5 x 32 2 1.999 79.92 lim x 1.9999 79.99 2 x4 2x3 32.) 1 5 1 1 5 2.0 ? 2x3 x 4x2 2.0001 80.01 4x2 2 8x 8x 16 2.001 80.08 16 80. 2.01 80.80 2.1 88.41 -4 -25 3 Analytically, lim x2 x2 x2 lim x4 (Hint: Use long division to factor x5 sin x 5x sin x x 1 5 1 2 67. lim x0 x0 lim 1 68. lim 31 x0 cos x x tan x0 lim 3 1 cos x x 3 0 0 69. lim x0 sin x 1 cos x 2x2 x0 lim sin x x 0 cos x x 70. lim cos 0 lim sin 0 1 1 1 0 2 sin2 x x0 x sin x sin x x 71. lim x0 lim 1 sin 0 0 72. lim tan2 x x0 x sin2 x 2 x0 x cos x lim 1 0 0 x0 lim sin x x sin x cos2 x 73. lim 1 h0 cos h h 2 h0 lim 1 cos h 1 h 0 cos h 74. lim sec 1 0 0 cos x cot x 75. lim x 2 x lim sin x 2 1 76. lim x 4 1 tan x sin x cos x cos x sin x sin x cos x cos2 x sin x cos x lim x 4 cos x sin x cos x 1 lim x 4 cos x x lim 4 77. lim t0 sin 3t 2t lim t0 sin 3t 3t 3 2 1 3 2 3 2 x lim 4 sec x 2 78. lim x0 sin 2x sin 3x x0 lim 2 sin 2x 2x 1 3 3x sin 3x 21 1 1 3 2 3 Section 1.3 sin 3t t 0.1 2.96 0.01 2.9996 0.001 3 sin 3t 3t 0 ? 0.001 3 0.01 2.9996 0.1 2.96 Evaluating Limits Analytically 63 4 79. f t t f t -2 -1 2 Analytically, lim sin 3t t0 t The limit appears to equal 3. 31 3. lim 3 t0 80. From the graph, lim x0 cos x 1 2x2 0.1 0.2498 0.25 - 1 x f x 1 0.2298 0.01 0.25 0.01 0.25 0.1 0.2498 1 0.2298 -1 x0 lim cos x 1 2x2 0.25 cos x cos x 1 1 cos2 x 1 2x2 cos x 1 2x2 sin2 x cos x 1 1 2 cos x 0.25 1 Analytically, cos x 1 2x2 sin2 x x2 x0 lim sin2 x x2 sin x2 x 1 2 cos x 1 1 1 4 1 4 1 81. f x x f x -2 2 0.1 0.099998 sin x2 x 0.01 0.01 lim x 0.001 0.001 sin x2 x2 0 ? 01 0.001 0.001 0. 0.01 0.01 0.1 0.099998 -1 Analytically, lim x0 x0 82. f x sin x 3 x -3 2 3 x f x 0.1 0.215 0.01 0.0464 lim 3 0.001 0.01 sin x x x 2 0 ? 0.001 0.01 0 1 0. 0.01 0.0464 0.1 0.215 -2 sin x Analytically, lim 3 x0 x f x x x f x The limit appear to equal 0. x0 83. lim x 0 lim 2x x x 0 3 x 2x 3 lim 2x 2 x x 0 3 x 2x 3 lim x 0 2 x x 2 64 Chapter 1 f x x x Limits and Their Properties f x x x x 4 x 4 x x x x x x x x x x x x 1 x x x x x x x 1 2 x x x 84. lim x0 lim x0 lim x0 lim x 0 lim x 0 85. lim f x x0 x x x x f x lim x x0 x x x 2 lim x0 4x x x 4x x xx x x2 4x lim 4 x xx 2x x x0 4 x2 x2 4x x 4 x x2 4x 86. lim f x f x x0 lim lim x x 2x 4x x 4 x0 lim 4 x2 x0 x x x2 x0 lim 2x x0 x 2x 4 87. lim 4 x0 x2 lim f x lim 4 x0 x0 88. lim b xa x a lim f x lim b xa xa x a 4 lim f x 4 x0 b lim f x b xa Therefore, lim f x x0 4. Therefore, lim f x xa b. 89. f x x cos x 4 90. f x x sin x 6 91. f x x sin x 6 -3 2 3 2 -2 -4 -2 -2 2 2 -6 x0 lim x cos x 0 x0 lim x sin x 0 x0 lim x sin x 0 92. f x x cos x 6 93. f x x sin 1 x 94. h x x cos 1 x 0.5 0.5 -2 2 -0.5 0.5 - 0.5 0.5 -6 -0.5 - 0.5 x0 lim x cos x 0 x0 lim x sin 1 x 0 x0 lim x cos 1 x 0 95. We say that two functions f and g agree at all but one point (on an open interval) if f x g x for all x in the interval except for x c, where c is in the interval. 97. An indeterminant form is obtained when evaluating a limit using direct substitution produces a meaningless fractional expression such as 0 0. That is, f x lim xc g x for which lim f x xc xc 96. f x x2 x except x 1 and g x 1 1. x 1 agree at all points 98. If a function f is squeezed between two functions h and g, h x f x g x , and h and g have the same limit L as x c, then lim f x exists and equals L. xc lim g x 0 Section 1.3 sin x x Evaluating Limits Analytically sin2 x x 65 99. f x x, g x 3 sin x, h x 100. f x x, g x 2 sin2 x, h x f g -5 h 5 -3 g 3 h f -3 -2 When you are "close to" 0 the magnitude of f is approximately equal to the magnitude of g. Thus, g f 1 when x is "close to" 0. When you are "close to" 0 the magnitude of g is "smaller" than the magnitude of f and the magnitude of g is approaching zero "faster" than the magnitude of f. Thus, g f 0 when x is "close to" 0. 101. s t lim t5 16t2 s5 5 st t 1000 lim t5 600 16t2 5 t 1000 lim t5 16 t t 5 t 5 5 lim t5 16 t 5 160 ft sec. Speed 160 ft sec 1000 16 0 5 10 seconds 2 1000 t 16 t lim t5 10 2 102. s t 16t2 s lim 1000 st 0 when t t5 10 2 5 10 2 5 10 2 lim t t5 10 2 16t2 5 10 2 125 2 t 16 t2 lim t5 10 2 5 10 2 t t 5 10 2 5 10 2 5 10 2 16 t lim t5 10 2 5 10 2 80 10 ft sec 253 ft sec 103. s t lim s3 t3 3 4.9t2 st t 150 lim t3 4.9 32 4.9 3 3 150 3 t 3 t 150 4.9 t 4.9t2 t lim t3 150 lim t3 4.9 9 t2 3 t 29.4 m sec lim t3 4.9 3 t 104. 4.9t2 150 0 when t a is 4.9a2 4.9 a 1500 49 5.53 seconds. The velocity at time t lim sa ta a st t lim ta 150 a t 4.9t2 t 2a 4.9 150 lim ta 4.9 a a t a t t lim ta 9.8a m sec. 54.2 m sec. Hence, if a 1500 49, the velocity is 9.8 1500 49 66 Chapter 1 Limits and Their Properties 1 x. lim f x and lim g x do not exist. x0 x0 105. Let f x lim f x 1 x and g x gx x0 1 lim x0 x 1 x x0 lim 0 0 106. Suppose, on the contrary, that lim g x exists. Then, since xc lim f x exists, so would lim f x g x , which is a xc xc contradiction. Hence, lim g x does not exist. xc 107. Given f x b, show that for every > 0 there exists a > 0 such that f x b < whenever x c < . Since f x b b b 0 < for any > 0, then any value of > 0 will work. 108. Given f x xc x n, n is a positive integer, then xc lim x n lim x x n xc 1 1 2 109. If b 0, then the property is true because both sides are equal to 0. If b 0, let > 0 be given. Since lim f x L, there exists > 0 such that 2 3 xc lim x lim x n xc xc xc c lim x x n xc xc c lim x lim xn . . . cn . c c lim xxn f x L < b whenever 0 < x wherever 0 < x c < , we have b f x L < xc c < . Hence, bL < or bf x bL. which implies that lim bf x 110. Given lim f x xc 0: 111. xc M f x lim M f x f xgx M f x lim f x g x lim M f x xc xc xc xc For every > 0, there exists > 0 such that f x 0 < whenever 0 < x c < . 0 f x f x 0 < for Now f x x c < . Therefore, lim f x 0. x c M 0 lim f x g x M 0 0 lim f x g x 0 Therefore, lim f x g x x c 0. 112. (a) If lim f x xc 0, then lim xc f x 0. (b) Given lim f x xc L: > 0 such that x c < . L < for f x xc f x f x lim f x lim f x xc xc xc lim f x 0 lim f x 0 Therefore, lim f x xc For every > 0, there exists f x L < whenever 0 < Since f x L f x x c < , then lim f x xc L. 0. 113. False. As x approaches 0 from the x left, 1. x 2 114. False. lim x sin x x 0 0 115. True -3 3 -2 116. False. Let f x x x 3 x 1 , 1 c 1. 1. 117. False. The limit does not exist. 4 118. False. Let f x 1 2 2x and g x x2. Then lim f x x1 1 but f 1 -3 6 Then f x < g x for all x 0. lim g x 0. But lim f x x0 x0 -2 Section 1.3 1 cos x x Evaluating Limits Analytically 1 cos x x cos2 x cos x 1 1 1 cos x cos x sin2 x x 0 x 1 cos x lim 67 119. Let f x lim f x 4, 4, if x 0 if x < 0 x0 120. lim x0 x0 lim 1 x0 x 1 lim lim sin x x x0 lim 4 4. x0 sin x cos x lim 1 sin x cos x x0 lim f x does not exist since for x < 0, f x x 0, f x 4. 4 and for x0 lim sin x x 0 x0 1 0 0, if x is rational 1, if x is irrational 0, if x is rational x, if x is irrational x0 121. f x g x 122. f x sec x x2 1 0, 2 n . (a) The domain of f is all x (b) -3 2 2 lim f x does not exist. 3 2 No matter how "close to" 0 x is, there are still an infinite number of rational and irrational numbers so that lim f x x0 does not exist. x0 -2 lim g x 0 When x is "close to" 0, both parts of the function are "close to" 0. The domain is not obvious. The hole at x apparent. (c) lim f x x0 0 is not 1 2 1 sec x x2 x2 1 sec x sec x 1 1 sec2 x x2 sec x 1 1 1 (d) sec x x2 tan2 x sec x 1 1 lim 1 sin2 x 1 2x cos x2 sec x 1 sin2 x 1 cos2 x x2 sec x 1 2 1 . 2 1 2 x 2 1 2 x for x 2 0. Hence, lim x0 sec x x2 x0 1 11 123. (a) lim 1 x0 cos x x2 x0 lim 1 cos x x2 1 1 cos x cos x (b) Thus, 1 cos x x2 1 1 2 cos x cos x 1 x0 lim 1 cos2 x x2 1 cos x sin2 x x2 1 2 1 2 1 1 cos x (c) cos 0.1 (d) cos 0.1 1 1 0.1 2 x0 lim 1 2 0.995 0.9950, which agrees with part (c). 124. The calculator was set in degree mode, instead of radian mode. 68 Chapter 1 Limits and Their Properties Section 1.4 1. (a) lim f x x3 Continuity and One-Sided Limits 1 1 1 2. (a) (b) x 2 lim f x lim f x 2 2 2 3. (a) lim f x x3 0 0 0 (b) lim f x x3 x 2 (b) lim f x x3 (c) lim f x x3 (c) lim f x x 2 (c) lim f x x3 The function is continuous at x 3. 4. (a) (b) lim f x lim f x 2 2 2 The function is continuous at x 2. 5. (a) lim f x 2 2 The function is NOT continuous at x 3. 6. (a) (b) lim f x lim f x 0 2 x 2 x4 x4 x 1 (b) lim f x x4 x 2 x 1 (c) lim f x x 2 (c) lim f x does not exist The function is NOT continuous at x 4. (c) lim f x does not exist. x 1 The function is NOT continuous at x 2. x x2 5 25 1 x 5 1 10 x x2 The function is NOT continuous at x 1. x 4 1 x x x x 1 x x x 2 2 2 2 2 4 x 4 4 x 1 4 1 4 x x 2 2 2 7. lim x5 x5 lim 8. lim x2 2 x2 x x x2 lim 9. x does not exist because x2 9 decreases without bound as x 3 . x 3 lim 9 10. lim x4 2 4 x4 lim x4 lim lim x4 11. lim x0 x x 1 x0 lim x x 1 x 1 12. lim x2 x x 2 2 x2 lim 1 13. lim x x0 x x x0 lim x xx x x x 1 x x0 lim x xx 1 xx 1 0 2x x x x x 1 1 2 x x 1 x2 x 1 x x0 lim xx x x 2 14. lim x x x x2 x x0 x0 lim x2 2 x x x x2 x x0 lim 2x x 2x 0 1 x x0 lim 2x x 2 2 5 2 2x 15. lim f x x3 x3 lim Section 1.4 16. lim f x x2 x2 Continuity and One-Sided Limits 69 x2 lim x2 4x 4x 6 2 2 2 17. lim f x x1 x1 x1 lim x 1 1 2 2 lim f x x2 lim x2 lim f x x1 lim x3 x2 lim f x 2 x1 lim f x 2 18. lim f x x1 x1 lim 1 x 0 19. lim cot x does not exist since x x 20. lim sec x does not exist since x 2 x 21. lim 3 x x4 5 33 5 4 lim cot x and lim cot x do not x lim 2 sec x and x lim 2 sec x do x 3 for 3 x < 4 exist. not exist. 22. lim 2x x2 x 22 2 2 23. lim 2 x3 x does not exist 24. lim 1 x1 x 2 1 1 2 because x3 lim 2 x 2 3 5 and x3 lim 2 x 2 4 6. 25. f x 1 x2 4 26. f x x2 x 1 1 27. f x x 2 x 2 and has discontinuities at x x 2 since f 2 and f 2 are not defined. x, 2, 2x 25 5, 5 . has a discontinuity at x 1 since f 1 is not defined. has discontinuities at each integer k since lim f x lim f x . xk xk 28. f x 1, x < 1 x 1 has a discontinuity at x x > 1 30. f t on 1 since f 1 2 x1 lim f x 1. 29. g x on x2 is continuous 3 3, 3 . 9 t2 is continuous 31. lim f x x0 3 x0 lim f x . 1, 4 . f is continuous on 1 x2 1 32. g 2 is not defined. g is continuous on 1, 2 . 33. f x x2 2x for all real x. 1 is continuous 34. f x real x. is continuous for all 35. f x 3x cos x is continuous for all real x. 36. f x cos x is continuous for all real x. 2 37. f x x x2 whereas x x x2 is not continuous at x 0, 1. Since 2 x x 1 is a nonremovable discontinuity. x x x 1 1 for x 0, x 0 is a removable discontinuity, 38. f x and x has nonremovable discontinuities at x 1 1 1 since lim f x and lim f x do not exist. x1 x 1 39. f x x x2 1 is continuous for all real x. 70 Chapter 1 Limits and Their Properties x 2 2 x 40. f x x x2 3 9 41. f x x 5 3 since has a nonremovable discontinuity at x lim f x does not exist, and has a removable x 3 discontinuity at x 3 since lim f x lim 1 x 3 1 . 6 has a nonremovable discontinuity at x 5 since lim f x x5 does not exist, and has a removable discontinuity at x 2 since lim f x lim 1 x 5 1 . 7 x3 x3 x 2 x 2 42. f x x x 1 2 x 1 43. f x x x 2 2 2 since has a nonremovable discontinuity at x 2 since lim f x does not exist, and has a removable x 2 discontinuity at x 1 since lim f x lim 1 x 2 1 . 3 has a nonremovable discontinuity at x lim f x does not exist. x 2 x1 x1 44. f x x x x, x2, 3 has a nonremovable discontinuity at x 3 x 1 x > 1 1. 3 since lim f x does not exist. x3 45. f x 46. f x 2x x2, 3, x < 1 x 1 1. has a possible discontinuity at x 1. f 1 2. x1 has a possible discontinuity at x 1. f 1 12 1 x1 1 x1 lim f x lim f x x1 lim x lim x2 1 1 x1 lim f x 1 2. x1 lim f x lim f x x1 lim lim x2 2x 1 3 1 lim f x x1 1 x1 x1 x1 x1 3. f 1 lim f x 1, therefore, f is continuous for 3. f 1 lim f x 1, therefore, f is continuous for f is continuous at x all real x. x 2 3 f is continuous at x all real x. 47. f x 1, x, x 2 x > 2 2. 48. f x x2 2x, 4x 1, x 2 x > 2 2. has a possible discontinuity at x 1. f 2 2 2 1 lim 2 x 2 1 x 1 2 has a possible discontinuity at x 1. f 2 2. x2 22 x2 4 2x x2 4x 4 1 lim f x does 3 not exist. x2 x2 lim f x x2 lim f x lim f x lim lim 2. x2 lim f x x2 lim 3 x2 lim f x does not exist. x2 x2 Therefore, f has a nonremovable discontinuity at x 2. Therefore, f has a nonremovable discontinuity at x 2. Section 1.4 x tan 4 , x, x tan 4 , x, Continuity and One-Sided Limits 71 x < 1 x 1 1 < x < 1 x 1 or x 1 1, x 1 1 x1 49. f x 50. f x csc x , 6 2, csc x , 6 2, x x 3 2 3 > 2 1 x 5 x < 1 or x > 5 1, x csc 5. 5 6 2 x5 has possible discontinuities at x 1. f 1 1 1 x 1 1. has possible discontinuities at x 1. f 1 2. lim f x x1 f 1 x1 2. lim f x x 1 lim f x csc 6 2 2 f 5 lim f x 2 3. f 1 lim f x f 1 lim f x x5 f is continuous at x real x. 1, therefore, f is continuous for all 3. f 1 x1 lim f x f 5 lim f x f is continuous at x 1 and x continuous for all real x. 52. f x 2k tan 5, therefore, f is csc 2x has nonremovable discontinuities at integer 51. f x multiples of 2. x has nonremovable discontinuities at each 2 1, k is an integer. x has nonremovable discontinuities at each 53. f x x integer k. 55. lim f x x0 x0 1 has nonremovable discontinuities at each 3 54. f x integer k. 56. lim f x x0 x0 0 0 2. -8 50 0 0 4. -8 20 lim f x lim f x f is not continuous at x 8 -10 f is not continuous at x 8 -10 57. f 2 8 8 a 8 22 2. 58. lim g(x x0 x0 lim 4 sin x x 2x 4 a Find a so that lim ax2 x2 x0 lim g x 4. 3a x0 lim a Let a 59. Find a and b such that lim a 3a 4a a b x2 xa x lim xa x 1 ax b a b 2 and lim ax x3 b b 2. b b 2 2 4 1 2 a2 a a 2a 8 a 4. 1 1 f x 2, x 2, 1, x 1 1 < x < 3 x 3 60. lim g x xa 61. f g x x 1 2 lim x Continuous for all real x. Find a such that 2a 72 Chapter 1 1 x Limits and Their Properties 1 5 1 6 x2 1 1 62. f g x 1 1. Continuous for all 63. f g x x2 Nonremovable discontinuity at x x > 1. 64. f g x sin x2 Nonremovable discontinuities at x Continuous for all real x 1 1 x 65. y x x 66. h x x 2 1 and x 2. Nonremovable discontinuity at each integer 0.5 -3 Nonremovable discontinuities at x 2 3 -3 -1.5 -2 4 67. f x 2x x2 4, 2x, x 3 x > 3 3 68. f x f 0 cos x x 5x, 50 0 lim 1, x < 0 x 0 -7 3 2 Nonremovable discontinuity at x 5 x0 -5 7 lim f x x0 cos x x 0 0 1 0 -3 x0 lim f x x0 x0 lim 5x Therefore, lim f x -5 f 0 and f is continuous on the entire real line. (x 0 was the only possible discontinuity.) 69. f x x x2 1 , 70. f x x x 3 3, Continuous on x 4 Continuous on 71. f x sec 72. f x x 1 x Continuous on: . . . , 6, 2 , sin x x Continuous on 0, 2, 2 , 2, 6 , 6, 10 , . . . x3 x 8 2 73. f x 3 14 74. f x 4 -4 -2 -4 0 4 The graph appears to be continuous on the interval 4, 4 . Since f 0 is not defined, we know that f has a discontinuity at x 0. This discontinuity is removable so it does not show up on the graph. The graph appears to be continuous on the interval 4, 4 . Since f 2 is not defined, we know that f has a discontinuity at x 2. This discontinuity is removable so it does not show up on the graph. Section 1.4 75. f x 1 4 16 x 33 16 Continuity and One-Sided Limits 3x 2 is continuous on 0, 1 . 2 73 x3 3 is continuous on 1, 2 . 76. f x f 0 x3 f 1 4. By the Intermediate Value and f 2 0 for at least one value of c between Theorem, f c 1 and 2. 2 and f 1 By the Intermediate Value Theorem, f x one value of c between 0 and 1. 4 x 4 0 for at least 77. f x x2 2 cos x is continuous on 0, . 78. f x f 1 tan tan x is continuous on 1, 3 . 8 8 < 0 and f 3 f 0 3 and f Value Theorem, f c between 0 and . 2 1 > 0. By the Intermediate 0 for the least one value of c 4 3 tan 3 > 0. 8 By the Intermediate Value Theorem, f 1 one value of c between 1 and 3. x3 80. f x x3 3x 2 0 for at least 79. f x x 1 f x is continuous on 0, 1 . f 0 1 and f 1 1 f x is continuous on 0, 1 . f 0 2 and f 1 2 0 for at least By the Intermediate Value Theorem, f x one value of c between 0 and 1. Using a graphing utility, we find that x 0.6823. 81. g t 2 cos t 3t By the Intermediate Value Theorem, f x 0 for at least one value of c between 0 and 1. Using a graphing utility, we find that x 0.5961. 82. h 1 3 tan g is continuous on 0, 1 . g0 2 > 0 and g 1 1.9 < 0. h is continuous on 0, 1 . h0 1 > 0 and h 1 2.67 < 0. By the Intermediate Value Theorem, g t 0 for at least one value c between 0 and 1. Using a graphing utility, we find that t 0.5636. x2 0 for at least By the Intermediate Value Theorem, h one value between 0 and 1. Using a graphing utility, we 0.4503. find that 84. f x x2 6x 8 83. f x x 1 f is continuous on 0, 5 . f 0 1 and f 5 1 < 11 < 29 The Intermediate Value Theorem applies. x2 x2 x x c x 4 x x 1 12 3 11 0 0 3 29 f is continuous on 0, 3 . f 0 8 and f 3 1 < 0 < 8 The Intermediate Value Theorem applies. x2 x x c 6x 2 x 8 4 0 0 4 1 2 or x 2 (x 0. 4 or x 3 (x 11. 4 is not in the interval.) 4 is not in the interval.) Thus, f 2 Thus, f 3 74 Chapter 1 Limits and Their Properties x2 x x 1 85. f x x3 x2 x 2 86. f x f is continuous on 0, 3 . f 0 2 and f 3 2 < 4 < 19 The Intermediate Value Theorem applies. x3 x3 x x2 x2 2 19 5 f is continuous on 2 , 4 . The nonremovable discontinuity, x 1, lies outside the interval. f x x 2 6 3 x 4 0 0 2 5 2 35 and f 4 6 20 3 20 35 < 6 < 6 3 The Intermediate Value Theorem applies. x2 x x2 x2 x x c 5x 2 x x 1 x 6 3 6 6x 0 0 3 6 2 x x (x2 x 3 has no real solution.) c 2 Thus, f 2 4. 2 or x 3 (x 2 is not in the interval.) 6. Thus, f 3 87. (a) The limit does not exist at x (b) The function is not defined at x c. c. (c) The limit exists at x c, but it is not equal to the value of the function at x c. (d) The limit does not exist at x c. 88. A discontinuity at x c is removable if you can define (or redefine) the function at x c in such a way that the new function is continuous at x c. Answers will vary. (a) f x (b) f x x 2 x 2 sin x 2 x 2 (c) f x 1, 0, 1, 0, y 3 2 1 if x 2 if 2 < x < 2 if x 2 if x < 2 -3 -2 -1 x -1 -2 -3 1 2 3 89. 5 4 3 2 1 -2 -1 -2 -3 y x 1 3 4 5 6 7 90. If f and g are continuous for all real x, then so is f g (Theorem 1.11, part 2). However, f g might not be continuous if g x 0. For example, let f x x and x2 1. Then f and g are continuous for all real x, gx but f g is not continuous at x 1. The function is not continuous at x lim f x 1 0 lim f x . x3 x3 3 because Section 1.4 91. True 1. f c xc Continuity and One-Sided Limits 75 92. True; if f x L is defined. L exists. gx,x c, then lim f x xc xc lim g x (if c. they exist) and at least one of these limits then does not equal the corresponding function value at x 2. lim f x 3. f c xc lim f x All of the conditions for continuity are met. 93. False; a rational function can be written as P x Q x where P and Q are polynomials of degree m and n, respectively. It can have, at most, n discontinuities. 95. lim f t t4 t4 94. False; f 1 is not defined and lim f x does not exist. x1 28 56 96. The functions agree for integer values of x: gx f x 3 3 x x 3 3 x x 3 x for x an integer lim f t At the end of day 3, the amount of chlorine in the pool has decreased to about 28 oz. At the beginning of day 4, more chlorine was added, and the amount was about 56 oz. However, for non-integer values of x, the functions differ by 1. f x 3 x 1 2 gx 3 1 0 2 3, g 1 2 x. 3 1 4. For example, f 1.04, 1.04 1.04 0 < t 2 t > 2, t is not an integer t > 2, t is an integer 97. C 0.36 t 0.36 t 1, 2, 98. N t t Nt 25 2 0 50 t 2 1 25 2 t 1.8 5 2 50 3 25 3.8 5 Nonremovable discontinuity at each integer greater than or equal to 2. You can also write C as C C 1.04, 1.04 0.36 2 t, 0 < t 2 . t > 2 Discontinuous at every positive even integer. The company replenishes its inventory every two months. N 50 Number of units 4 3 2 1 40 30 20 10 t 2 t 1 2 3 4 4 6 8 10 12 Time (in months) 0 (t 0 corresponds to 8:00 A.M., s 20 99. Let s t be the position function for the run up to the campsite. s 0 k, r 10 0. Let f t to campsite)). Let r t be the position function for the run back down the mountain: r 0 When t When t 0 (8:00 A.M.), f 0 s0 s 10 r0 0 k < 0. k (distance st rt. 10 (8:10 A.M.), f 10 r 10 > 0. 0. If f t 0, then Since f 0 < 0 and f 10 > 0, then there must be a value t in the interval 0, 10 such that f t st rt 0, which gives us s t r t . Therefore, at some time t, where 0 t 10, the position functions for the run up and the run down are equal. 76 Chapter 1 Limits and Their Properties 101. Suppose there exists x1 in a, b such that f x1 > 0 and there exists x2 in a, b such that f x2 < 0. Then by the Intermediate Value Theorem, f x must equal zero for some value of x in x1, x2 (or x2, x1 if x2 < x1). Thus, f would have a zero in a, b , which is a contradiction. Therefore, f x > 0 for all x in a, b or f x < 0 for all x in a, b . 100. Let V 4 r 3 be the volume of a sphere of radius r. V is 3 continuous on 1, 5 . V1 V5 4 3 4 3 4.19 53 523.6 Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 such that V r 275. (In fact, r 4.0341.) 102. Let c be any real number. Then lim f x does not exist xc since there are both rational and irrational numbers arbitrarily close to c. Therefore, f is not continuous at c. 103. If x 0, then f 0 0 and lim f x x0 continuous at x 0. If x t x 0. Hence, f is 0, then lim f t tx t x 0 for x rational, whereas 0 for x irrational. Hence, f is 0. lim f t lim kt kx not continuous for all x y 104. sgn x 1, if x < 0 0, if x 0 1, if x > 0 1 1 -4 -3 -2 -1 4 3 2 1 x 1 2 3 4 105. (a) 60 50 40 30 20 10 S (a) lim sgn x x0 -2 -3 -4 (b) lim sgn x x0 x0 t 5 10 15 20 25 30 (c) lim sgn x does not exist. (b) There appears to be a limiting speed and a possible cause is air resistance. 0 b y 106. (a) f x 0 x < b b < x 2b (b) g x x 2 b y 0 x b x 2 b < x 2b 2b b 2b x b 2b b x NOT continuous at x b. b 2b Continuous on 0, 2b . 1 x2, x c x, x > c 107. f x f is continuous for x < c and for x > c. At x c, you need 1 c2 c. Solving c2 c 1, you obtain c 1 2 1 4 1 2 5 . 108. Let y be a real number. If y 0, then x 0. If y > 0, then let 0 < x0 < 2 such that M tan x0 > y (this is possible since the tangent function increases without bound on 0, 2 ). By the Intermediate Value Theorem, f x tan x is continuous on 0, x0 and 0 < y < M , which implies that there exists x between 0 and x0 such that tan x y. The argument is similar if y < 0. Section 1.4 x c2 x c Continuity and One-Sided Limits 77 109. f x , c > 0 c2 and x x x x x c2 x c2 c 2 Domain: x lim x c2 0 x c2 x c lim 0, c c2 c x x c2, 0 c2 c2 x c c 0, x0 x0 x0 lim x0 lim 1 c2 c 1 2c Define f 0 1 2c to make f continuous at x 0. 110. 1. f c is defined. 2. lim f x xc 111. h x x f c exists. x0 -3 xx 15 lim f c x0 Let x 3. lim f x xc c f c. x. As x c, 3 -3 Therefore, f is continuous at x c. h has nonremovable discontinuities at x 112. (a) Define f x f a f2 a f2 x 1, 2, 3, . . . . f1 x . Since f1 and f2 are continuous on a, b , so is f. f2 b f1 b < 0 0. f1 a > 0 and f b 0 f1 c By the Intermediate Value Theorem, there exists c in a, b such that f c f c f2 c f1 c x and f2 x f2 c 2 , f1 0 < f2 0 and f1 cos c . (b) Let f1 x cos x, continuous on 0, 2 > f2 2. Hence by part (a), there exists c in 0, Using a graphing utility, c 113. The statement is true. If y 0 and y 1, then y y Case 1: If x y y y 1 2, 2 such that c 0.739. 1 0 x2, as desired. So assume y > 1. There are now two cases. 1 2y and 1 1 2 then 2x yy x Case 2: If x y 1 2 1 2 2 1 2y 2y 1 2y 2y x2 y y2 > y2 y y 1 1 4 x2 x2 2x 2y y y x In both cases, y y 2 1 x2. 114. P 1 P2 P5 P 02 P 12 P 22 1 1 1 P0 P1 P2 2 2 2 1 1 1 1 2 5 x for infinitely many values of x. Continuing this pattern, we see that P x x for all x. Hence, the finite degree polynomial must be constant: P x 78 Chapter 1 Limits and Their Properties Section 1.5 1. lim 2 2 Infinite Limits 2. lim 1 x 1 x 2 x 2 x x2 x x2 4 4 x x 2 2 3. x 2 lim tan x 4 x 4 4. x 2 lim sec x 4 x 4 x 2 lim 2 x 2 lim lim tan x 2 lim sec 5. f x 1 x2 9 3.5 0.308 3.1 1.639 3.01 16.64 3.001 166.6 2.999 166.7 2.99 16.69 2.9 1.695 2.5 0.364 x f x x 3 lim f x x 3 lim f x 6. f x x x2 9 3.5 1.077 3.1 5.082 3.01 50.08 3.001 500.1 2.999 499.9 2.99 49.92 2.9 4.915 2.5 0.9091 x f x x 3 lim f x x 3 lim f x 7. f x x2 x2 9 3.5 3.769 3.1 15.75 3.01 150.8 3.001 1501 2.999 1499 2.99 149.3 2.9 14.25 2.5 2.273 x f x x 3 lim f x x 3 lim f x 8. f x sec x 6 3.5 3.864 3.1 19.11 3.01 191.0 3.001 1910 2.999 1910 2.99 191.0 2.9 19.11 2.5 3.864 x f x x 3 lim f x x 3 lim f x Section 1.5 1 x2 1 x2 4 x 4 x 2 3 Infinite Limits 79 9. lim x0 x0 lim 10. lim x2 2 3 Therefore, x 0 is a vertical asymptote. x2 lim Therefore, x x2 2 x 2 x 1 x x2 2 2 x 1 2 is a vertical asymptote. 2 is a vertical asymptote. 11. lim x2 12. lim x0 2 x2 1 x x x0 lim 2 x2 1 x x x2 lim Therefore, x x1 0 is a vertical asymptote. Therefore, x lim x lim 2 x2 1 2 x2 1 x x x x 1 is a vertical asymptote. x 1 x2 2 2 x 1 x1 lim x 1 lim x2 2 x 2 x 1 1 is a vertical asymptote. x2 x2 4 Therefore, x Therefore, x x2 x2 4 13. x 2 lim and lim x 2 Therefore, x x2 2 is a vertical asymptote. and lim x2 lim x2 x 2 x2 x 2 4 4 Therefore, x 2 is a vertical asymptote. 15. No vertical asymptote since the denominator is never zero. sin 2x has vertical asymptotes at cos 2x 1 4 4 n , n any integer. 2 14. No vertical asymptote since the denominator is never zero. 16. s 5 lim h s and lim h s s 5 . 17. f x x tan 2x 2n Therefore, s s5 5 is a vertical asymptote. and lim h s s5 lim h s . Therefore, s 5 is a vertical asymptote. 1 has vertical asymptotes at cos x 1 , n any integer. 4 t2 4 t2 18. f x x sec x 2n 2 19. lim 1 t0 t0 lim 1 Therefore, t 0 is a vertical asymptote. 20. g x 1 2 x3 x2 4x 3x2 6x 24 1 x, 6 x 2, 4 1 x x2 6 x2 2x 2x 8 8 No vertical asymptotes. The graph has holes at x 2 and x 4. 80 Chapter 1 x 2 x x 2 x Limits and Their Properties 21. x 2 lim x x 1 1 2 is a vertical asymptote. x 2 lim Therefore, x lim lim x x x1 x 2 x x 2 x 1 1 x1 Therefore, x 1 is a vertical asymptote. 4 x2 x 6 2x2 9x 18 0 and x 1 4x xx 3 x 2 x2 2 9 4 xx 3 22. f x x x3 ,x 3, 2 3 and x 2. x2 2x2 4 x x x 2 x 2 x2 2 1 Vertical asymptotes at x x3 x 1 1 x 3. The graph has holes at x 23. f x 1 x2 x x 1 24. h x x3 2 has no vertical asymptote since x 1 has no vertical asymptote since 3. x 2 lim f x x 1 lim x2 x 1 1. lim h x x 2 lim The graph has a hole at x x x2 2 1 2. 4 . 5 The graph has a hole at x x x 5 x 5 x2 3 1 x x2 3 ,x 1 tt 2 t 2 2 t2 25. f x 5 5. 26. h t t 4 t t 2 t2 4 ,t 2 No vertical asymptotes. The graph has a hole at x Vertical asymptote at t t 2. tan 2n 2 1 2 sin cos 2. The graph has a hole at 27. s t t has vertical asymptotes at t sin t n ,n 28. g has vertical asymptotes at n , n any integer. 0 since a nonzero integer. There is no vertical asymptote at t 0 since lim t0 t sin t 1. There is no vertical asymptote at lim tan 1. 0 29. lim x x2 1 x 1 1 2 x 1 lim x 1 2 30. lim x2 x 6x 1 2 7 x 1 x 1 lim x 7 8 -3 -3 3 3 -5 -12 Removable discontinuity at x 1 Removable discontinuity at x 1 Section 1.5 x2 x x2 x 1 1 1 1 1 -3 Infinite Limits 81 31. x 1 lim 8 32. lim 3 x 1 sin x 1 x 1 1 -3 2 x 1 lim Removable discontinuity at x 1 3 Vertical asymptote at x -8 -2 33. lim x2 x x 3 2 x2 3 x 34. lim x1 2 1 x x x2 1 2 35. lim x3 x x2 x2 3 36. lim x4 x2 16 37. x 3 lim 2x x 6x2 4x2 3 6 x 4x x 3 lim x x 1 x 2 x 3x 2x 3 3 x 3 lim x x 1 2 4 5 3x 2x 1 3 x x2 2 sin x 5 8 2 1 9 2 cos x 38. x lim 1 2 1 3 x lim 1 2 1 2x 3 2x 1 1 x lim 1 2 39. lim x1 x 2 x2 x 1 x 1 x 1 x1 lim x x2 1 1 2 1 x 40. lim x3 41. lim x0 1 42. lim x0 x2 43. lim x0 44. x lim 2 45. lim x x csc x x lim x sin x 0 46. lim x0 x 2 cot x x0 lim x 2 tan x 0 47. x 1 2 lim x sec x and x 1 2 lim x sec x . 48. x 1 2 lim x2 tan x x 1 2 and x2 x 1 2 lim x2 tan x . Therefore, lim x sec x 1 2 x does not exist. Therefore, lim tan x does not exist. 49. f x lim f x x2 x3 x 1 lim 1 x 1 x 1 x2 x 1 x2 1 x 1 50. f x lim f x x3 x2 x x1 4 1 1 lim x x 1 1 x2 x2 x 0 x 1 1 x1 x1 3 x1 -4 5 -8 8 -3 -4 51. f x 1 x2 25 -8 0.3 52. f x 8 sec x 6 -9 6 x5 lim f x x3 -0.3 lim f x 9 -6 82 Chapter 1 Limits and Their Properties 54. The line x c is a vertical asymptote if the graph of f approaches as x approaches c. 53. A limit in which f x increases or decreases without bound as x approaches c is called an infinite limit. is not a number. Rather, the symbol xc lim f x says how the limit fails to exist. x 3 6 x x 2 x2 4x 3 12 1 x2 1 55. One answer is f x x . 56. No. For example, f x vertical asymptote. has no 57. 3 2 1 -2 -1 -1 -2 y 58. P lim k V 59. (a) r (b) r (c) 50 sec2 50 sec2 lim 6 3 200 ft sec 3 200 ft sec k k V (In this case we know that k > 0.) V0 x 1 3 2 50 sec2 60. C 528x , 0 x < 100 100 x $176 million $528 million (c) C 75 (d) x100 (a) C 25 (b) C 50 $1584 million 528 100 x . Thus, it is not possible. lim 61. m lim m m0 1 v2 c2 vc 62. (a) r m0 (b) r (c) lim 27 625 49 2 15 625 225 2x 625 x2 7 ft sec 12 3 ft sec 2 vc lim 1 v2 c2 x25 63. (a) Average speed 50 50 50y 50x 50x 50x x 25x 25 Total distance Total time 2d d x 2xy y 2xy 2xy 2y x y 50y 25 x d y (b) x y 30 150 25x 25 40 6.667 50 50 60 42.857 (c) lim x25 x As x gets close to 25 mph, y becomes larger and larger. Domain: x > 25 Section 1.5 64. (a) Infinite Limits 83 x f x 1 0.1585 0.5 0.5 0.0411 0.2 0.0067 0.1 0.0017 x 0.01 0 0.001 0 0.0001 0 x0 -1.5 1.5 lim sin x x 0 -0.25 (b) x f x 1 0.1585 0.25 0.5 0.0823 0.2 0.0333 0.1 0.0167 0.01 0.0017 0.001 0 0.0001 0 -1.5 1.5 x0 lim x sin x x2 0 -0.25 (c) x f x 1 0.1585 0.25 0.5 0.1646 0.2 0.1663 0.1 0.1666 0.01 0.001 0.0001 0.1667 0.1667 0.1667 -1.5 1.5 x0 lim x sin x x3 0.1667 1 6 -0.25 (d) x f x 1 0.1585 1.5 0.5 0.3292 0.2 0.8317 0.1 1.6658 0.01 16.67 0.001 166.7 0.0001 1667.0 -1.5 1.5 x0 lim x sin x x4 -1.5 For n > 3, lim x0 x sin x xn . 65. (a) A 1 bh 2 1 2 r 2 1 10 10 tan 2 50 tan 50 1 10 2 2 (b) f 0.3 0.47 0.6 4.21 0.9 18.0 1.2 68.6 1.5 630.1 Domain: (c) 100 0, 2 (d) lim A 2 0 0 1.5 84 Chapter 1 Limits and Their Properties (b) The direction of rotation is reversed. (d) L (e) 450 66. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 1700 2 850 revolutions per minute. (c) 2 20 cot 2 10 cot : straight sections. The angle subtended in each circle is 2 2 2 2 . 0.3 306.2 0.6 217.9 0.9 195.9 1.2 189.6 1.5 188.5 Thus, the length of the belt around the pulleys is 20 Total length Domain: 0, 2 2 10 60 cot 2 30 30 2 2 . (f) 0 0 2 lim 2 L 60 188.5 (All the belts are around pulleys.) (g) lim L 0 67. False; for instance, let f x x2 x 1 . 1 68. False; for instance, let f x gx x2 x x x2 1 1 or 1 . 69. True The graph of f has a hole at 1, 2 , not a vertical asymptote. 70. False; let f x 1, x 3, x x 0 0. 0, but 71. Let f x lim 1 x2 1 x2 1 and g x x2 and lim 1 x4 1 x4 x2 1 , and c x4 , but 1 x4 0. x0 x0 The graph of f has a vertical asymptote at x f 0 3. x0 lim x0 lim 0. 72. Given lim f x xc and lim g x xc L: (2) Product: If L > 0, then for L 2 > 0 there exists 1 > 0 such that g x L < L 2 whenever 0 < x c < 1. Thus, then for M > 0, there exists 2 > 0 such that f x > M 2 L whenever L 2 < g x < 3L 2. Since lim f x xc x c < 2. Let be the smaller of 1 and 2. Then for 0 < x c < , we have f x g x > M 2 L L 2 M. Therefore lim f x g x . The proof is similar for L < 0. xc (3) Quotient: Let There exists 0 < x gx f x 1 > 0 be given. > 0 such that f x > 3L 2 whenever 0 < x c < 1 and there exists 2 > 0 such that g x 1 and L < L 2 whenever 0 < x c < 2. This inequality gives us L 2 < g x < 3L 2. Let be the smaller of 2. Then for c < , we have < 3L 2 3L 2 gx f x 0. . Therefore, lim xc 73. Given lim f x x c , let g x 1. then lim x c gx fx 0 by Theorem 1.15. Review Exercises for Chapter 1 85 74. Given lim Then, lim xc 1 f x 0. Suppose lim f x exists and equals L. xc xc xc 75. f x 1 x 3 is defined for all x > 3. Let M > 0 be > 0 such that f x 1 xc f x lim 1 lim f x 1 L 0. given. We need 1 x 3 > M whenever 3 < x < 3 Equivalently, x So take 1 x 1 > 3 8 . 3 < , x > 3. 3 < , This is not possible. Thus, lim f x does not exist. xc 1 3 < whenever x M 1 . Then for x > 3 and x M M and hence f x > M. 76. f x 1 x 4 is defined for all x < 4. Let N < 0 be given. We need 4 > 1 whenever x N 4 < , x < 4. Equivalently, 4 < > 0 such that f x 1 x 4 < N whenever 4 < x < 4. Equivalently, x 1 x x 4 1 1 . Note that N > 0 because N < 0. For x and x < 4, 1 whenever x 4 < , x < 4. So take N 1 1 1 > N, and < N. 4 x 4 x 4 < Review Exercises for Chapter 1 1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3. Or, the length is slightly longer than the distance between the two points, approximately 8.25. 4 3. f x x f x x0 2. Precalculus. L 9 1 2 3 1 2 8.25 x 2 x 0.1 2 1 -3 3 0.01 1.0050 0.001 1.0005 0.001 0.9995 0.01 0.9950 0.1 -5 1.0526 1.0 0.9524 lim f x 4. 3 x f x lim f x 0.1 1.432 1.414 0.01 1.416 0.001 1.414 0.001 1.414 0.01 1.413 0.1 1.397 -3 3 x0 -3 5. h x x2 x 2x (a) lim h x x0 2 3 6. g x 3x x 2 (a) lim g x does not exist. x2 (b) lim h x x 1 (b) lim g x x0 0 7. lim 3 x1 x 3 1 2 . Then for 0 < x 1 < , you have Let > 0 be given. Choose x 1 3 x f x 1 < x < 2 < L < . 86 Chapter 1 x 9 Limits and Their Properties 3 8. lim Let x9 > 0 be given. We need x 3 < x 3 x 3 < 5 . x 3 x 9 < x 3. Assuming 4 < x < 16, you can choose Hence, for 0 < x x x f x 9 < x 5 , you have 3 9 < 5 < 3 < L < . 9. lim x2 x2 3 1 3 1 < x2 4 x 2 x 2 < 2 < Let > 0 be given. We need x2 x 2 < 1 x 2 . Assuming, 1 < x < 3, you can choose x x 2 x x2 x 2 5. Hence, for 0 < x 5 you have 2 < 2 < 4 < 1 < 5 < 1 x 2 3 f x L < . > 0 be given. 10. lim 9 x5 9. Let can be any positive 11. lim t4 t 2 4 2 6 2.45 number. Hence, for 0 < x 9 f x 9 < L < . 5 < , you have 12. lim 3 y y4 1 34 1 9 13. lim t t 2 2 t 2 4 t 2 lim 1 t 2 4 1 4 x x 2 14. lim t3 t2 t 4 x 1 4 9 3 x lim t t3 3 6 15. lim x x x4 2 4 x4 lim x 1 x x 2 2 1 2 x 2 1 4 1 1 1 1 s0 16. lim x0 x0 lim 2 1 4 4 4 x x 2 2 x4 lim 1 4 2 x xx 1 1 1 s 1 1 s s 1 1 s 5 x2 x 5x 5x 5 25 s 1 x0 lim x 2 17. lim 1 x x0 1 x 1 s s 1 x0 lim x0 lim x 1 18. lim 1 1 s0 s0 lim 1 1 1 25 75 1 1 1 s s s 1 1 1 1 1 s x2 2 x3 1 4 8 1 2 x x x x2 2x 2 x 2 2 x2 2x 4 2 4 4 12 1 3 s0 lim lim 19. lim x x3 5 x 125 5 x 5 lim x 20. lim x x 2 lim x 5 lim x2 x 2 lim Review Exercises for Chapter 1 1 cos x sin x 6 x x sin x 1 2 1 cos x x sin 1 2 87 21. lim x0 x0 lim 1 0 0 22. x lim 4 4x tan x 1 2 4 1 4 23. lim sin x x0 lim lim 6 cos x cos x x sin x 1 1 x0 cos x lim x0 6 sin x 3 2 sin x x x0 0 3 1 2 3 2 24. lim cos x x 1 x0 lim lim 0 cos cos x cos x x sin x 1 sin x x x0 x0 lim sin x0 0 1 1 2 0 26. lim f x xc 25. lim f x xc gx 3 4 2 3 2g x 3 4 2 2 3 7 12 27. f x (a) x f x lim 2x x 1 1 1.1 0.5680 3 1.01 0.5764 3 1 3 1 0.577 lim 1.001 0.5773 1.0001 0.5773 3 3. 2x 2x 3 1 1 3 3 (b) 2 2x x 2x x 1 1 x1 Actual limit is 2x x 2x 1 1 1 3 2x 1 3 3 3 3 -1 0 2 (c) lim x1 x1 x1 lim lim x 2x 1 2 1 x1 2 2 3 1 x x f x lim 1 x 1 x 3 3 1 3 28. f x (a) x 1 1.1 0.3228 x 1 1.01 0.3322 0.333 lim lim lim 1 x 3 1.001 0.3332 1.0001 0.3333 1 3. 3 3 (b) -3 2 3 x1 Actual limit is x 1 1 1 1 3 3 3 -3 3 (c) lim x1 x 1 x1 x x 3 x 2 x 2 x1 x 1 1 1 1 3 x x x 2 x2 1 3 x1 x 3 88 Chapter 1 sa ta a st t Limits and Their Properties 4.9 4 4.9 t 2 29. lim lim t4 200 4 4 4.9t2 t 200 30. s t 0 t2 4.9t2 200 0 6.39 sec lim t4 4 t 4 t 4 40.816 t st t When a 39.2 m sec t6.39 6.39, the velocity is approximately sa a t6.39 lim t4 4.9 t lim lim 4.9 a t 6.39 4.9 6.39 62.6 m sec. 31. lim x3 x x 3 3 x3 lim 1 x x 3 3 32. lim x x4 1 does not exist. The 4. 33. lim f x x2 0 graph jumps from 2 to 3 at x 34. lim g x x1 1 1 2 35. lim h t does not exist because t1 t1 t1 36. lim f s s 2 2 lim h t 1 1 2 1 1 1 2 and 1. lim h t 37. f x xk xk x 3 3 3 k k 3 where k is an integer. 2 where k is an integer. 38. f x 3x2 x x 1 2 3x x ,1 lim x lim x 2 x 1 1, . 1 f is continuous on Nonremovable discontinuity at each integer k Continuous on k, k 1 for all integers k 3x2 x x1 39. f x lim f x x 1 2 2 3x x 5 2 x 1 1 40. f x lim 5 5 x, 2x 3, x 3 3 1 x 2 x > 2 x1 lim 3x x2 Removable discontinuity at x ,1 1, Continuous on 1 x2 lim 2x Nonremovable discontinuity at x Continuous on ,2 2, 2 41. f x lim 1 x 1 x 2 2 2 2 42. f x lim 1 x x 1 x , 1 1 1 x x2 x0 Nonremovable discontinuity at x Continuous on ,2 2, 2 Domain: 1 , 0, 0 Nonremovable discontinuity at x , 1 0, Continuous on 43. f x x1 x1 3 x 1 44. f x lim x 2x x 2x 1 2 1 1 1 2 1 1, lim f x lim f x 1 x 1 Nonremovable discontinuity at x , 1 1, Continuous on Removable discontinuity at x , 1 Continuous on Review Exercises for Chapter 1 x 2 89 45. f x csc 46. f x tan 2x 47. f 2 5 x2 Nonremovable discontinuities at each even integer. Continuous on 2k, 2k 2 Nonremovable discontinuities when x 2n 4 1 Find c so that lim cx c2 6 2c 5 1 1 2 6 5. Continuous on 2n 4 1 , 2n 4 1 c for all integers k. for all integers n. 48. lim x x1 x3 1 1 2 4 x1 lim x Find b and c so that lim x2 Consequently we get Solving simultaneously, 1 b b bx c c 2 2 and lim x2 x3 bx c c c 4. 4. 50. C 4. and 9 3 and 3b 1 < 0 and 49. f is continuous on 1, 2 . f 1 f 2 13 > 0. Therefore by the Intermediate Value Theorem, there is at least one value c in 1, 2 such that 2c3 3 0. 9.80 9.80 2.50 2.50 x x 1 1, x > 0 30 C has a nonremovable discontinuity at each integer. 0 0 5 51. f x x2 x x2 4 2 x 4 4 2 x x 2 2 52. f x x 1x ,0 0 0 1, (a) lim f x (b) lim f x x2 x2 (a) Domain: (b) lim f x x0 (c) lim f x x1 (c) lim f x does not exist. 53. g x 1 2 x 0 54. h x 4x 4 x2 2 and 55. f x 8 x 10 2 Vertical asymptote at x Vertical asymptotes at x x 2 Vertical asymptote at x 10 56. f x csc x 57. x 2 lim 2x2 x x 2 1 Vertical asymptote at every integer k x 2x 1 58. x 1 2 lim 59. x 1 lim x x3 x2 x 1 1 x 1 lim x2 1 x 1 1 3 60. x 1 lim x x4 1 1 x 1 lim x 2 1 1 x 1 1 4 61. lim x1 2x 1 1 90 Chapter 1 x2 x 2x 1 Limits and Their Properties 1 1 x3 sec x x cos2 x x 1 x2 62. x 1 lim 63. lim x x0 64. lim x2 3 4 65. lim x0 sin 4x 5x csc 2x x x0 lim 4 sin 4x 5 4x 1 x sin 2x 4 5 66. lim x0 67. lim x0 x 0 lim 68. lim x0 69. C 80,000p , 0 p < 100 100 p $14,117.65 $720,000 (b) C 50 (d) lim $80.000 80,000p 100 p (a) C 15 (c) C 90 p100 70. f x (a) x tan 2x x 0.1 2.0271 2 0.01 2.0003 0.001 2.0000 0.001 2.0000 0.01 2.0003 0.1 2.0271 f x lim tan 2x x x0 (b) Yes, define f x tan 2x , x 2, x x 0 . 0 0. Now f x is continuous at x Problem Solving for Chapter 1 1. (a) Perimeter PAO x2 x2 Perimeter PBO x x (b) r x x2 x x2 12 4 33.02 33.77 0.98 1 1 0 0 1 1 12 x4 2 9.08 9.60 0.95 2 2 1 y x2 1 1 2 2 1 2 2 x2 x2 x2 x2 1 1 y2 x4 y2 x4 1 1 1 1 2. (a) Area Area (b) a x PAO PBO Area Area 1 bh 2 1 bh 2 PBO PAO 4 2 1 2 2 0 1 1 x 2 1 1 y 2 x2 2 x 2 1 1 2 1 2 1 x x 2 y 2 x2 2 1 y2 x4 x2 x2 1 x4 x4 x x Perimeter PAO Perimeter PBO rx (c) lim r x x0 0.1 1 20 1 200 1 10 0.01 1 200 1 20,000 1 100 0.1 2.10 2.00 1.05 0.01 2.01 Area Area PAO PBO 2 8 4 3.41 3.41 1 ax 2.00 1.005 (c) lim a x x0 x0 lim x Problem Solving for Chapter 1 4 3 3 4 0 0 4 3 Tangent line: y 4 y h = sin 1 91 3. (a) There are 6 triangles, each with a central angle of 60 3. Hence, Area hexagon 6 1 bh 2 6 1 1 sin 2 3 4. (a) Slope (b) Slope 3 x 4 3 x 4 3 25 4 3 3 2 h = sin 60 1 60 2.598. (c) Let Q mx (d) lim mx x3 x, y 25 x x3 x, x2 3 4 25 x x x 25 25 3 3 3 3 25 x2 lim x2 3 x2 25 4 16 x2 x x2 4 25 25 4 4 6 4 x2 x2 4 4 Error: 3 3 2 0.5435 x3 lim (b) There are n triangles, each with central angle of 2 n. Hence, An (c) n An 1 n bh 2 6 2.598 1 2 n 1 sin 2 n 12 3 24 3.106 48 3.133 n sin 2 2 96 3.139 n . x3 lim x 3 25 x x2 4 x3 lim 3 4 This is the slope of the tangent line at P. (d) As n gets larger and larger, 2 Letting x An 2 n, sin 2 n 2 n . n approaches 0. sin 2 n 2 n sin x x which approaches 1 12 5 5 . 12 5. (a) Slope 6. a bx x 3 a a a bx x bx bx 3 3 3 a a bx bx 3 3 (b) Slope of tangent line is y 12 y (c) Q mx (d) lim mx x5 x Letting a Thus, 5 x 12 5 x 12 x, 169 x x5 5 169 Tangent line 12 169 x2 5 12 169 x 5 x2 12 12 169 169 x2 x2 x2 3 simplifies the numerator. x0 lim 3 bx x 3 x0 lim x 3 3 bx bx b bx 6. 3 . 3 x, y x0 lim lim 12 Setting Thus, a b 3 3 3 and b 3, you obtain b 6. x5 lim lim x x 12 144 169 x2 5 12 169 x2 x2 5 12 x 5 169 25 169 x2 12 x2 10 12 5 12 x5 x5 lim This is the same slope as part (b). 92 Chapter 1 Limits and Their Properties 3 2 28 12 -0.1 7. (a) 3 x1 x1 3 3 0 (b) 3 27 27, x 3 x x x1 x2 1 3 -30 0.5 (c) x 27 lim f x 27 1 3 27 1 1 14 2 x Domain: x 0.0714 1 x1 3 (d) lim f x x1 x1 lim 2 3 1 x1 3 1 x2 x1 3 3 3 4 3 x1 x1 2 3 3 2 2 x1 lim lim 3 1 x1 3 x1 x x1 1 1 1 3 3 1 1 x1 3 3 2 x1 3 2 x1 lim 3 3 1 12 1 lim a2 1 1 2 a2 2 8. lim f x x0 x0 x0 2 2 x0 9. (a) lim f x x2 3: g1, g4 lim f x x0 lim ax tan x a because lim tan x x 1 (b) f continuous at 2: g1 (c) lim f x x2 3: g1, g3, g4 Thus, a2 a2 a a 2 2 1 a y 3 2 1 -1 x -1 -2 1 a 0 0 1, 2 11. 4 3 2 1 -4 -3 -2 -1 -2 -3 -4 x 1 2 3 4 2 a 10. y (a) f 1 4 4 1 3 4 0 1 1 0 (b) f (a) f 1 f 0 f 1 2 f 3 f 1 (b) x1 x1 x0 x0 1 0 0 3 1 1 1 1 2 1 1 1 0 1 lim f x 1 1 lim f x lim f x 2.7 x1 lim f x lim f x lim f x x1 (c) f is continuous for all real numbers except x 0, 1, 1, 1, . . . 2 3 x1 2 lim f x (c) f is continuous for all real numbers except x 0, 1, 2, 3, . . . Problem Solving for Chapter 1 192,000 r v2 v02 93 12. (a) v2 192,000 r r lim r v02 48 48 13. (a) 2 y 192,000 v2 v02 48 192,000 48 v02 48 1920 r v2 v02 4 3 mi sec. 1 x a b v0 Let v0 (b) v2 1920 r r lim r (b) (i) lim Pa, b x xa 1 0 0 1 v02 2.17 2.17 (ii) lim Pa, b x xa (iii) lim Pa, b x xb (iv) lim Pa, b x xb v2 1920 v02 2.17 (c) Pa, b is continuous for all positive real numbers except x a, b. (d) The area under the graph of U, and above the x-axis, is 1. 1.47 mi sec . v0 1920 2.17 v02 2.17 mi sec v2 10,600 v02 6.99 Let v0 (c) r lim r v0 10,600 6.99 v02 6.99 2.64 mi sec. Let v0 Since this is smaller than the escape velocity for Earth, the mass is less. 14. Let a Let 0 and let > 0 be given. There exists 0 < 1 a . Then for 0 < x x < ax < f ax 1 1 1 > 0 such that if 0 < x 0 < 1 then f x L < . a , you have a 1 L < . 0 and f x x0 x0 As a counterexample, let a Then lim f x x0 1, 2, lim f 0 x x 0 . 0 x0 1 L, but lim f ax lim 2 2. ...
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This note was uploaded on 09/01/2008 for the course BIOL 1362 taught by Professor Richardknapp during the Spring '08 term at Spring Hill.

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