Trascendentes Tempranas Solucionario- Dennis Zill - 4th Edition - Capitulo 2

www.elsolucionario.org Chapter 2 Limit of a Function 2.1 Limits — An Informal Approach ππ ππ 1. lim2 2(3x + 2) = 8 2. lim2 2(x2 − 1) = 3 x→2 x→2 8 • (2, 8) 6 4 (–2, 3) • 3 –π –π -4 -2 2 4 -4 4 ππ π π√ 3. No2 limit 2 as x → 0. 4. lim2 2 x − 1 = 2 x→5 4 5 (5, 2) • –π -4 -2 2 4 –π -5 5 -5 -4 π πx − 1 π πx − 3x 2 2 5. lim2 2 = lim (x + 1) = 2 6. lim2 2 = lim (x − 3) = −3 x→1 x − 1 x→1 x→0 x x→0 5 5 (1, 2) –π -5 5 –π -5 5 (0, 3) -5 -5 73 74 CHAPTER 2. LIMIT OF A FUNCTION 7. No limit as x → 3. 8. No limit as x → 0. 3 3 3 6 -3 3 -3 -3 x3 x4 − 1 9. lim =0 10. lim =2 x→0 x x→1 x2 − 1 3 3 (1, 2) -3 (0, 0) 3 -3 3 -3 -3 ππ 11. lim f (x) = 3 12. No2 limit 2 as x → 2. x→0 (0, 3) • 3 3 -3 3 –π -3 3 -3 -3 13. lim f (x) = 0 14. No limit as x → 0. x→2 3 3 (2, 0) -3 3 -3 3 -3 -3 15. (a) 1 (b) -1 (c) 2 (d) doesn’t exist 16. (a) 0 (b) 3 (c) 3 (d) 3 17. (a) 2 (b) -1 (c) -1 (d) -1 18. (a) doesn’t exist (b) 3 (c) -2 (d) doesn’t exist 19. Correct √ 20. Incorrect; lim+ 4 x=0 x→0 2.1. LIMITS — AN INFORMAL APPROACH 75 √ 21. Incorrect; lim− 1−x=0 x→1 22. Correct 23. Incorrect; lim bxc = 0 x→0+ 24. Correct 25. Correct 26. Incorrect; lim cos−1 x = 0 x→0− √ 27. Incorrect; lim 9 − x2 = 0 x→3− 28. Correct 29. (a) Does not exist (b) 0 (c) 3 (d) −2 (e) 0 (f) 1 30. (a) ≈ 2.5 (b) 1 (c) −1 (d) Does not exist (e) 0 (f) 0 ππ ππ 22 22 31. 4 32. 4 2 –π -4 4 –π -4 -2 2 4 -2 -4 -4 ππ ππ 22 22 33. 4 34. 4 2 2 –π -4 -2 2 4 –π -4 -2 2 4 -2 -2 -4 -4 ππ ππ 22 22 35. 1 36. 0.5 0.5 –π -1 -0.5 0.5 1 –π -0.5 0.5 -0.5 -0.5 -1 The limit does not exist. The limit is 0. www.elsolucionario.org 76 CHAPTER 2. LIMIT OF A FUNCTION ππ ππ 22 22 37. 0.5 38. -0.5 0.5 -1 –π -0.5 0.5 –π -2 -0.5 -3 The π π limit is −0.25. The π π limit is −3. 22 22 39. 40. -0.5 0.5 5 -1 –π –π -0.5 0.5 -2 -5 -3 The limit is −2. The limit does not exist. 41. x→1 − 0.9 0.99 0.999 0.9999 f (x) −3.25536642 −3.02276607 −3.00225263 −3.00022503 x → 1+ 1.1 1.01 1.001 1.0001 f (x) −2.79817601 −2.97775903 −2.99775260 −2.99977503 lim f (x) = −3 x→1 42. x → 1− 0.9 0.99 0.999 0.9999 f (x) 1.05360516 1.00503359 1.00050033 1.00005000 x → 1+ 1.1 1.01 1.001 1.0001 f (x) 0.95310180 0.99503309 0.99950033 0.99995000 lim f (x) = 1 x→1 43. x → 0− −0.1 −0.01 −0.001 −0.0001 f (x) −0.04995835 −0.00499996 −0.00050000 −0.00005000 x → 0+ 0.1 0.01 0.001 0.0001 f (x) 0.04995835 0.00499996 0.00050000 0.00005000 lim f (x) = 0 x→0 1 − cos x 44. Since is an even function, it suffices to consider only x → 0+ . x2 x → 0+ 0.1 0.01 0.001 0.0001 f (x) 0.49958347 0.49999583 0.49999996 0.50000000 lim f (x) = 0.5 x→0 x 45. Since is an even function, it suffices to consider only x → 0+ . sin 3x x → 0+ 0.1 0.01 0.001 0.0001 f (x) 0.33838634 0.33338334 0.33333383 0.33333334 4 .00334672 1.99 −1.0001 f (x) 5.00000033 1.61000000 12.999 0.0001 f (x) 0. x → 4− 3.49792633 −0.48008703 −0.49979176 −0. x → 3− 2.9 3.41000000 11.25 x→4 48.0001 f (x) −0.94010000 11.999 −1. 15 2.50002083 x → 3+ 3.00600400 5.9999 f (x) 11.0001 f (x) 12.99400400 4.9999 f (x) 4.24984395 0. Since is an even function.99 2.2.50020843 −0.25158234 0. x x → 0+ 0.99940001 lim f (x) = 12 x→−2 2.1 3.2.00003333 1.01 3.99 3.1 1. x → −2− −2.5 x→3 49.06040010 5.00000000 lim f (x) = 1 x→0 47.94039900 4.1 0.999 2.99 0.01 0.00060004 lim f (x) = 5 x→1 50. −3 5.06010000 12.25000156 x → 4+ 4.9 −1. LIMIT THEOREMS 77 lim f (x) = 0.01 1.00060001 x → −2+ −1. it suffices to consider only x → 0+ .2 Limit Theorems 1.64100000 5.25015645 0.9999 f (x) 0.001 −2.999 3. −12 4.9 2.24999844 lim f (x) = 0.52186477 −0.24845673 0.1 −2.01 −2.43900000 4.001 0.0001 f (x) 1.33333333 x→0 tan x 46.1 4.99940004 x → 1+ 1.001 4.001 3.9999 f (x) −0.001 1.00600100 12.49997917 lim f (x) = −0.99400100 11. x → 1− 0. cos π = −1 3.01 4.24998438 0.9 0.25001563 0.50209311 −0. 5 x→1. 3 19. 14 12.5)(x + 3) 28.5 . −136 9. 4 13. 7 18. −1 16. 16 y 2 − 25 21. lim = lim (y − 5) = −10 y→−5 y + 5 y→−5 u2 − 5u − 24 22.5 x − 1. lim = lim = lim = x→6 x2 − 7x + 6 x→6 (x − 1)(x − 6) x→6 x − 1 5 15. LIMIT OF A FUNCTION 6. 16 √ 17. lim = lim = lim x(x + 5) = 14 x→2 x−2 x→2 x−2 x→2 2x2 + 3x − 9 (2x − 3)(x + 3) 2(x − 1.5 x→1. lim = lim (u + 3) = 11 u→8 u−8 u→8 x3 − 1 (x − 1)(x2 + x + 1) 23. does not exist 11.78 CHAPTER 2.5 x − 1. lim = lim = lim =− x→−3 4x − 36 2 x→−3 4(x + 3)(x − 3) x→−3 2(x − 3) 12 x3 + 3x2 − 10x x(x + 5)(x − 2) 27. 28/9 x2 − 6x x(x − 6) x 6 14. lim = lim = lim = lim 2(x + 3) = 9 x→1. 4 8. lim = lim = lim =− t→−1 t − 1 2 t→−1 (t + 1)(t − 1) t→−1 t−1 2 (x − 2)(x + 5) 8(15) 25. −125 7. −8/5 10. lim = = 60 x→10 x−8 2 2x + 6 2(x + 3) 1 1 26.5 x→1. does not exist 20.5 x − 1. lim = lim = lim (x2 + x + 1) = 3 x→1 x − 1 x→1 x−1 x→1 t3 + 1 (t + 1)(t2 − t + 1) t2 − t + 1 3 24. √3 4 r  2 2 r h h − 16 h 128 39. −210 or −1024 37. LIMIT THEOREMS 79 t3 − 2t + 1 (t − 1)(t2 + t − 1) t2 + t − 1 1 29. lim = lim = lim (16 + h) = 16 h→0 h h→0 h h→0 1 46. 16 r r 5 x3 − 64x 5 x2 − 64 41. lim x x + 4 3 x − 6 = −2 2 3 −8 = 4 2 x→−2   x2 + 3x − 1 1 x2 + 3x 33. lim [(1 + h)3 − 1] = lim (h2 + 3h + 3) = 3 h→0 h h→0 . −100. www. lim − 2 = lim − x→2 x − 2 x + 2x − 8 x→2 x − 2 (x − 2)(x + 4)   x+4 6 = lim − x→2 (x − 2)(x + 4) (x − 2)(x + 4) x−2 1 1 = lim = lim = x→2 (x − 2)(x + 4) x→2 x + 4 6 35. 000 43. lim− = lim− = −2 x→0 x2 + 2x x→0 x+2 42.org 2. does not exist 36. a2 − 2ab + b2 √ √ p 44. lim u2 x2 + 2xu + 1 = lim u2 − 2u + 1 = lim (u − 1)2 = |u − 1| x→−1 x→−1 x→−1 (8 + h)2 − 64 16h + h2 45. 2 √ 2 2 38.elsolucionario.2. lim = lim = lim = t→1 t + t − 2 3 2 t→1 (t − 1)(t + 2t + 2) 2 t→1 t + 2t + 2 2 5 x3 1 1 30. lim = lim (h2 + 8h + 16) = h→4 h+5 h−4 h→4 h+5 3 40. lim x3 (x4 + 2x3 )−1 = lim = lim = x→0 x→0 x4 + 2x3 x→0 x + 2 2 (x + 2)(x5 − 1)3 2(−1) 1 31. lim+ √ = =− x→0 ( x + 4) 2 16 8 √ √ √ √ √ 32. lim + = lim = lim (x + 3) = 3 x→0 x x x→0 x x→0     1 6 1 6 34. 2 r 4 √ 56. 64 1 55. 2 . 8a 3 60.80 CHAPTER 2. lim = lim √ √ h→0 h h→0 h x+h+ x (x + h) − x 1 1 = lim √ √ = lim √ √ = √ h→0 h( x + h + x) h→0 x + h + x 2 x √ √ √ t−1 t−1 t+1 1 1 49. 8 59. LIMIT OF A FUNCTION     1 1 1 1 x − (x + h) −h 47. lim = lim √ = lim √ = t→1 t − 1 t→1 t − 1 t + 1 t→1 t + 1 2 √ √ √ u+4−3 u+4−3 u+4+3 50. lim = lim √ x→1 x −1 2 x→1 x2 − 1 4 + x + 15 1−x = lim √ x→1 (x + 1)(x − 1)(4 + x + 15) −(x − 1) = lim √ x→1 (x + 1)(x − 1)(4 + x + 15) −1 1 = lim √ =− x→1 (x + 1)(4 + x + 15) 16 53. does not exist 58. lim √ = lim √ √ √ v→0 1+v−1 v→0 1+v−1 1+v+1 25 + v + 5 √ v 1+v+1 1 = lim √ = v→0 v 25 + v + 5 5 √ √ √ 4 − x + 15 4 − x + 15 4 + x + 15 52. 32 54. lim − = lim = lim h→0 h x+h x h→0 h (x + h)x h→0 hx(x + h) 1 1 = lim − 2 =− 2 h→0 x + hx x √ √ √ √ √ √ x+h− x x+h− x x+h+ x 48. = 2 2 57. lim = lim √ u→5 u−5 u→5 u−5 u+4+3 u−5 1 1 = lim √ = lim √ = u→5 (u − 5)( u + 4 + 3) u→5 u+4+3 6 √ √ √  √ 25 + v − 5 25 + v − 5 25 + v + 5 1+v+1 51. 000 x→1 (x − 1)2 x→1 x − 1 x−1 2x 1 lim 1 62. (a) lim = 2 lim   = 2 · x→0 =2 x→0 sin x x→0 sin x sin x lim x x→0 x 1 − cos2 x sin2 x sin x sin x (b) lim = lim = lim · =1·1=1 x→0 x2 x→0 x2 x→0 x x   8x2 − sin x sin x sin x (c) lim = lim 8x − = lim 8x − lim = −1 x→0 x x→0 x x→0 x→0 x   sin x sin x 63. . Discontinuous at −3 and + nπ. Discontinuous at −1 and 1 nπ 5. . lim [2f (x) − 5] = lim (x + 3) = 5 · 4 = 20 x→2 x→2 x+3 2 lim f (x) − lim 5 = 20 x→2 x→2 20 + 5 lim f (x) = = 12. for n an integer 2 7. Continuous everywhere 2. CONTINUITY 81 x100 − 1 x100 − 1 61. Discontinuous at 2 8. 2 π 6. Discontinuous at 3 and 6 4. Discontinuous at . Discontinuous at 0 . 1. for n = 0.5 x→2 2 2. (a) lim = lim x→1 x2 − 1 x→1 (x + 1)(x − 1) 1 x100 − 1 1 = lim · = · 100 = 50 x→1 x + 1 x−1 2 x50 − 1 x50 − 1 x50 + 1 (b) lim = lim · 50 x→1 x − 1 x→1 x − 1 x +1 x100 − 1 1 1 = lim · 50 = 100 · = 50 x→1 x − 1 x +1 2 (x100 − 1)2 x100 − 1 x100 − 1 (c) lim = lim · = 100 · 100 = 10.3 Continuity 1. Continuous everywhere 3. .3.2. lim sin x = lim x · = lim x · lim =0·1=0 x→0 x→0 x x→0 x→0 x   2f (x) − 5 64. 2. lim+ f (x) = 3. 2 23. so m = 1 and x→3 x→3 n = 3. Discontinuous at 0 13. so m = 10/3 and n = −5/3. x→4 x→4 (x − 2)(x + 2) 26. and f (3) = n. Since lim− f (x) = 4m and lim+ f (x) = 16. www. Discontinuous at x < 0 and 2 11.org 82 CHAPTER 2. lim f (x) = 2m + n. π ]. 3π 2 ]. so x = 2π 3 + 2nπ or x = 4π 3 + 2nπ. Since sin x1 is discontinuous only at x = 0. (a) yes (b) yes 17. 24. Since lim− f (x) = 3m. it is continuous on [2. it is continuous on [ π1 . LIMIT OF A FUNCTION 9. f (x) is discontinuous on (−∞. Continuous everywhere 1 10. 4].elsolucionario. Thus. 22. we have 4m = 16 and m = 4. it is discontinuous on [−1. we obtain 3m = 10. Since f (x) is discontinuous only at x = 2. Discontinuous at e−2 12. (a) yes (b) no 20. Since f (x) is defined and continuous exactly on (1. 28. (a) no (b) no 18. . 25. 4] and discontin- uous on [1. x→2 x→2 x−2 27. (a) yes (b) yes 19. (a) yes (b) yes 14. Solving 2 + sec x = 0. (a) no (b) no 21. 5]. we have m − n = 5 and x→1− x→1+ 2m + n = 5. (a) yes (b) yes 16. ∞) and on [ π2 . (a) no (b) yes 15. Since lim f (x) = lim = 4 we have f (2) = m and m = 4. Adding. Since lim f (x) = m − n. 3] and continuous on (2. and f (1) = 5. we have 3m = 3 = n. ∞) and discontinuous on [ −2 π . 5]. we obtain cos x = − 12 . Since lim = lim = lim (x2 + 1) = 2. Discontinuous at . lim [1 + cos(cos x)] = 1 + cos lim cos x = 1 + cos(cos ) = 1 + cos 0 = 2 x→π/2 x→π/2 2     t2 − π 2 (t − π)(t + π) 37. lim tan 2 = tan lim = tan lim = tan = 3 t→0 t + 3t t→0 t(t + 3) t→0 t + 3 3 √ √ √ 39. define f (1) = 2. lim2 cos x = cos lim2 x = cos π = −1 x→π x→π   π 35. lim t − π + cos2 t = cos2 π = 1 = 1 t→π h i3 40. n an integer 2 3 -3 3 -3 30. lim cos = cos lim = cos 2π = 1 t→π t−π t→π t−π       πt πt π π √ 38. Discontinuous at every integer 3 -3 3 -3 √ √ x−9 ( x + 3)( x − 3) √ 31. lim (4t + sin 2πt)3 = lim (4t + sin 2πt) = (4 + sin 2π)3 = 43 = 64 t→1 t→1 . CONTINUITY 83 n 29.3. Since lim √ = lim √ = lim ( x + 3) = 6. x→1 x − 1 2 x→1 x2 − 1 x→1   √ π π 2π 3 33. define f (9) = 6.2. x→9 x − 3 x→9 x−3 x→9 x4 − 1 (x2 + 1)(x2 − 1) 32. lim sin(cos x) = sin lim cos x = sin(cos ) = sin 0 = 0 x→π/2 x→π/2 2   π 36. lim sin(2x + ) = sin lim (2x + ) = sin = x→π/6 3 x→π/6 3 3 2   √ √ 34. On [0. since −3 ≤ 1 ≤ 5. c = . since −1 ≤ 8 ≤ 15. On [−2. Since f (b) < g(b). 50. 3) ∪ (3. 2 2 47. f (0) = 10. f ◦ g is continuous for x + 3 > 0 or on (−3.84 CHAPTER 2. ∞). 5]. 4. there exists c ∈ (a. 5] such that c2 − 2c = 8. 1) ∪ (1. then (f − g)(a) > 0. f (−2) = 3. x+3 5(x − 2)2 5(x − 2)2 5(x − 2)2 44. 48. ∞). c = . Setting c2 + 1 = or c2 − = 0 gives us (c + )(c − ) = 0 c +1 4 4 2 2 1 1 or c = ± . On [−2. On [1. f (5) = 15. f (1) = −1. f (−2) = −3. f (c) = 0 for some c between 0 and 1. f (3) = 242. Setting c3 − 2c = 0 gives us c(c2 − 2) = 0. by the x+3 x−4 3 4 2 3 corollary to the Intermediate Value Theorem. since 3 ≤ 6 ≤ p13. 1] such that 2 = 8. 1]. then by the Intermediate Value Theorem f (c) = 0 for some c ∈ (0. ± 2. Since f (a) > g(a).√ 2] such that c3 − 2x + 1 = 1. c = 4. lim sin −1 = sin−1 lim x→−3 x2 + 4x + 3 x→−3 x2 + 4x + 3   x+3 = sin−1 lim x→−3 (x + 3)(x + 1)     1 1 π = sin−1 lim = sin−1 − =− x→−3 x + 1 2 6 lim cos 3x 42. 51. lim ecos 3x = ex→π = ecos 3π = e−1 x→π 1 43. c = 0. By the Intermediate Value Theorem. b) such that (f − g)(c) = 0. Since (f ◦ g)(x) = √ . Thus. then (f − g)(b) < 0. f (3) = 13. there exists c ∈ [1. x2 + 1 x4 + 1 1 1 1 2 52. 1). there exists 10 5 1 1 1 c ∈ [0. there exists −1 ± 1 − 4(1)(−5) c ∈ [−2. and −7 ≤ 50 ≤ 242. Since (f ◦g)(x) = = = . 1) if f (x) = 2x7 + x − 1 is 0 on (0. 1). 46. LIMIT OF A FUNCTION     x+3 x+3 41. f (1) = 5. we see that f ◦g is continuous (x − 2)2 − 1 x2 − 4x + 3 (x − 1)(x − 3) for x 6= 1 and x 6= 3 or on (∞. 2 2 49. By the Intermediate Value Theorem. Then f (0) = − > 0 and f (1) = − < 0. 2]. 3] such that f (c) = 50. By the corollary to the Intermediate Value Theorem. Setting c2 − 2c − 8 = 0 gives us (c − 4)(c + 2) = 0 or c = −2. since 5 ≤ 8 ≤ 10. then by the Intermediate Value Theorem there exists c ∈ [0. Since f (0) = −7. and hence between −3 and 4. The equation will have a solution on (0. 3] such that c2 + c + 1 = 6. 3]. f (2) = 5. . Setting c2 + c − 5 = 0 gives us c = = √ √ 2 −1 ± 21 −1 + 21 . Let f (x) = + . Then f (c) = g(c). 45. By the Intermediate Value Theorem. there exists c ∈ [−2. By the Intermediate Value Theorem. Since f (0) = −1 and f (1) = 2. 2] the zero is approximately 1.3.75. 1].org 2. 0] the zero is approx- imately −0. In [−1. 2).21. In the solution of Problem 52 we saw that there is a zero in [0. The bisection method gives 1800 r ≈ 3. Applying the bisection method on this interval. Since  π  2 = 2 . Solving the latter equation for h. 4] and [14. In [0. It is easily seen that the expression on the left side of this equation is negative when x = 2 and positive when x = 3. -3 3 -3 3 56. we find V = Sr − πr3 or 2πr3 − Sr + 2V = 0. by the corollary to the Intermediate Value Theorem.  π sin 54.84 ft 2πr 1800 and h = − r ≈ 4. −1] the zero is approximately −1. and 0 ≤ 1 ≤ 2 . 2 x 2 3 55.48 ft and r ≈ 14. The corresponding values of h are h = − r ≈ 78. π π 2 π 2 π sin x 1 there exists c between and π such that = .64. 2πr . 60. sin π = 0.78. 1] the zero is approximately 0. CONTINUITY 85 53. we observe zeros in [3. www. Let f (x) = e−x − ln x. In [−2. 2πr 1 Substituting into the formula for V .91 ft. -3 3 -3 57. 1]. 2 (b) 5000 10 20 -5000 (c) From the graph. we find c ≈ 0. Thus. then by the Intermediate Value Theorem. we find c ≈ 0. Applying the bisection method on [2. In [1. then the volume is given by V = πr2 h and the surface S area is S = 2πr2 + 2πrh. We want to solve f (x) = x5 + 2x − 7 = 50 or x5 + 2x − 57 = 0. we obtain h = − r. 3]. Then f (1) = e−1 − ln 1 = e−1 > 0 and f (2) = e−2 − ln2 < 0.21.elsolucionario. f (c) = 0 for some c ∈ (1. (a) If h is the height of the cylinder. Applying the bisection method to f (x) = 2x7 + x − 1 on [0. 58.75.34. 15]. 59. we find c ≈ 2.29 ft. n an integer. x ≥ 0 3 -3 3 -3 (g ◦ f )(x) = |x| − 1 is continuous at x = 0. f + g is continuous at a.4 Trigonometric Limits sin 3t 1 sin 3t 3 1. LIMIT OF A FUNCTION 61. we x→a x→a get: lim (f + g)(x) = lim [f (x) + g(x)] = lim f (x) + lim g(x) x→a x→a x→a x→a = f (a) + g(a) = (f + g)(a) Thus. In fact. or 3π/2. Therefore. x < 0 64. f ◦ g will be discontinuous for x = nπ/2. (f ◦ g)(x) = is continuous at x = 0. 2. In the interval [0. lim = lim = t→0 2t 2 t→0 t 2 . then lim f (x) = f (a) and lim g(x) = g(a). 63. 3 -3 3 -3 65. (a) For any real a. f /g is continuous at a. 2π). there are infinitely many points of the graph on the line y = 1 and infinitely many points of the graph on the line y = 0. Since f and g are continuous at a. the Dirichlet function is discontinuous at every real number. If x is irrational. f ◦ g will be discontinuous whenever cos x is an integer. lim f (x) does not exist since f takes on the values 0 and 1 arbitrarily x→a close to any real number.86 CHAPTER 2. 62. then lim f (x) = f (a) and lim g(x) = g(a).  |x + 1|. If x is rational. (b) The graph consists of infinitely many points on each of the lines y = 0 and y = 1. π/2. From this. Since f and g are continuous at a. (c) Let r be a positive rational number. this will be the case whenever x = 0. π. Thus. then x + r is rational so that f (x+r) = 1 = f (x). then x+r is irrational so that f (x+r) = 0 = f (x). From this we x→a x→a get: lim (f /g)(x) = lim [f (x)/g(x)] = lim f (x)/ lim g(x) x→a x→a x→a x→a = f (a)/g(a) = (f /g)(a) (since g(a) 6= 0) Thus. |x − 1|. between any two real numbers. lim = lim = 62 = 36 t→0 t2 t→0 t      t3 t2 1 1 12. θ→π/2 cos θ cos(3x − π/2) sin 3x 17. lim = = x→0 1 + cos x 1+1 2 cos 2x 5. lim = =0 x→0 4 + cos x 4+1 1 + sin x 1+0 1 4. TRIGONOMETRIC LIMITS 87 sin(−4t) 2. lim does not exist. lim = lim · t→0 sin t t→0 t sin t    sin(t/2) sin(t/2) 1 0 = lim lim = · =0 t→0 t t→0 (sin t)/t 2 1   2 sin2 6t sin 6t 11. lim does not exist. lim = lim = x→1 2x − 2 2 x→1 x − 1 2 x − 2π x − 2π 1 14. lim =1 x→0 cos 3x   tan x 1 sin x 1 1 1 6. lim (5t cot 2t) = 5 lim = 5 lim cos 2t · t→0 t→0 sin 2t t→0 (sin 2t)/t   1  1 5 = 5 lim cos 2t lim =5·1· = t→0 t→0 (sin 2t)/t 2 2   2 sin2 t sin t sin t 9. lim = lim · cos t = 4 · 1 = 4 t→0 t sec t csc 4t t→0 t   t cos 2t 1 8.4. lim = −4 t→0 t sin x 0 3. lim = 2 lim · =2·1·0=0 t→0 t cos2 t t→0 t cos2 t   sin2 (t/2) sin(t/2) t sin(t/2) 10. lim = lim · = (1 · 1) = x→0 3x 3 x→0 x cos x 3 3   1 sin 4t 7. lim = lim = lim =1 x→2π sin x x→2π sin(x − 2π) x→2π sin(x − 2π)/(x − 2π) cos x 15. x→0 x 1 + sin θ 16. lim = lim t · = lim t lim =0· 2 =0 t→0 sin2 3t t→0 sin2 3t t→0 t→0 [(sin 3t)/t]2 3 sin(x − 1) 1 sin(x − 1) 1 13.2. lim = lim =3 x→0 x x→0 x . org 88 CHAPTER 2. lim = lim · = lim lim =3· = t→0 sin 7t t→0 t sin 7t t→0 t t→0 (sin 7t)/t 7 7   sin 2t sin 2t t 20.elsolucionario. lim √ = lim t· = lim t lim =0·1=0 t→0 + t t→0 + t t→0 + t→0 + t √ √ 1 − cos t 1 − cos u 22. x2 + 2x − 8 x→2 x→2 (x − 2)(x + 4)   sin u 1 1 1 Letting u = x − 2. lim sin 2t csc 3t = lim = lim · t→0 t→0 sin 3t t→0 t sin 3t    sin 2t 1 1 2 = lim lim =2· = t→0 t t→0 (sin 3t)/t 3 3      sin t √ sin t √ sin t 21. lim = lim lim =1· = x→0 cos2 x − 1 x→0 cos x − 1 x→0 cos x + 1 2 2    sin x + tan x sin x sin x 1 28. lim = =1 cos 8t t→0 1 √ √ (x + 2 sin x)2 x2 + 4x sin x + 4 sin x 25. lim = lim 1 − 5 = 1 − 5 = −4 t→0 t2 t→0 t cos 4t 1 24. u→0 u u+6 6 6 . lim = lim + · = 1 + (1 · 1) = 2 x→0 x x→0 x x cos x sin 5x2 sin 5u 29. lim+ = lim+ x→0 x x→0 x   √ 4 sin x = lim+ x + 4 sin x + =0+0+4=4 x→0 x (1 − cos x)2 h i 1 − cos x  26. lim = lim · = lim · (1 + cos t) t→0 1 − cos t t→0 1 − cos t 1 + cos t t→0 sin2 t  2 h i 1 = lim lim (1 + cos t) = 12 · 2 = 2 t→0 (sin t)/t t→0 sin(x − 2) sin(x − 2) 31. we have lim √ = lim = 0. t→0 t u→0 u    t2 − 5t sin t sin t 23. First. lim = lim (1 − cos x) lim =0·0=0 x→0 x x→0 x→0 x    cos x − 1 cos x − 1 1 1 1 27. www. x→0 x2 u→0 u    2  t2 t2 1 + cos t t 30. lim = · · lim = lim = x→−2 4x + 8 5 4 x→−2 x+2 4 x→−2 5x + 10 4      sin 3t sin 3t t sin 3t 1 1 3 19. Letting u = x2 . we get lim · =1· = . Letting u = t. we have lim = lim = 5. rewrite lim as lim . LIMIT OF A FUNCTION sin(5x + 10) 5 1 sin(5x + 10) 5 sin(5x + 10) 5 18. Letting u = x − 3: x→3 sin(x − 3) x→3 sin(x − 3) h u i   1 1 lim · (u + 6) = lim · (u + 6) = · 6 = 6 u→0 sin u u→0 (sin u)/u 1   2 sin 4x + 1 − cos x 2 sin 4x 1 − cos x 33. we multiply out and simplify: (cos x − sin x)(1 + tan x) = cos x + cos x tan x − sin x − sin x tan x    cos x    sin x sin x = cos x + cos x − sin x − sin x cos x cos x cos x sin2 x cos2 x − sin2 x = cos x − = cos x cos x Substituting this result back into the function. lim = lim + =8+0=8 x→0 x x→0 x x   4x2 − 2 sin x 2 sin x 34. we have: 1 − tan x 1 1 √ lim = lim =√ = 2 x→π/4 cos x − sin x x→π/4 cos x 2/2 36. lim = lim 4x − = 0 − 2 = −2 x→0 x x→0 x 1 + tan x 35. TRIGONOMETRIC LIMITS 89 x2 − 9 (x − 3)(x + 3) 32.4. producing: 1 + tan x   1 − tan x 1 − tan x 1 + tan x lim = lim · x→π/4 cos x − sin x x→π/4 cos x − sin x 1 + tan x 1 − tan2 x = lim x→π/4 (cos x − sin x)(1 + tan x) Focusing first on the denominator. First. Using the trigonometric identity cos 2x = cos2 x − sin2 x. rewrite lim as lim . we get:   1 − tan2 x cos x = (1 − tan2 x) (cos x − sin x)(1 + tan x) cos2 x − sin2 x  2   2  sin x sin x cos x − cos x cos x − cos2 x cos x = = cos x − sin x 2 2 cos x − sin2 x 2  2 2   cos x − sin x 1 1 = = cos x cos x − sin x 2 2 cos x Finally. we have: cos 2x cos2 x − sin2 x lim = lim x→π/4 cos x − sin x x→π/4 cos x − sin x (cos x + sin x)(cos x − sin x) = lim x→π/4 cos x − sin x √ = lim (cos x + sin x) = 2 x→π/4 .2. returning to the limit. Start by multiplying the function by . lim 4 4 = lim 4 4 h→0 h h→0 h sin(π/4) cos h + cos(π/4) sin h − sin(π/4) = lim h→0 h √ √ √ ( 2/2) cos h + ( 2/2) sin h − ( 2/2) = lim h→0 h √ √   √ 2 cos h + sin h − 1 2 cos h − 1 sin h 2 = lim = lim + = 2 h→0 h 2 h→0 h h 2 π  π π  π f +h −f cos + h − cos 38. lim f (x) = 1. Since −1 ≤ cos ≤ 1. then x x x→0 x→0 π by the Squeeze Theorem. x→0 43. Since lim (−Bx2 ) = 0 and lim Bx2 = 0. However. since x2 ≥ 0 for all x. lim x cos = 0. x→2 x→2 x→2 44. For both limits. Thus. then by the Squeeze x→0 x→0 Theorem. Since lim −x2 = 0 and lim x2 = 0. x→0 x π π 40. lim x sin = 0. −Bx2 ≤ x2 f (x) ≤ Bx2 in that interval. LIMIT OF A FUNCTION π  π π  π f +h −f sin + h − sin 37. Similarly.90 CHAPTER 2. lim f (x) = 3. lim 6 6 = lim 6 6 h→0 h h→0 h cos(π/6) cos h − sin(π/6) sin h − cos(π/6) = lim h→0 h √ √ ( 3/2) cos h − (1/2) sin h − ( 3/2) = lim h→0 h "√    # √ 3 cos h − 1 1 sin h 3 1 1 = lim − = ·0− ·1=− h→0 2 h 2 h 2 2 2 1 1 39. Since −1 ≤ sin ≤ 1. Since lim (−x2 + 1) = 1 and lim (x2 + 1) = 1. we use the result from Problem 39. we have −x2 ≤ f (x) − 1. Since lim (2x − 1) = 3 and lim (x2 − 2x + 3) = 3. then f (x) − 1 ≤ x2 . x→0 . then −x2 ≤ x2 cos ≤ x2 . Since lim (−|x|) = 0 and lim |x| = 0. or −x2 + 1 ≤ f (x) for all x. then by the Squeeze Theorem. lim x sin = 0: x→0 x   1 1 1 (a) lim x3 sin = lim x2 · x sin = lim x2 · lim x sin = 0 · 0 = 0 x→0 x x→0 x x→0 x→0 x    1 1 1 (b) lim x2 sin2 = lim x sin lim x sin =0·0=0 x→0 x x→0 x x→0 x 42. |f (x)| ≤ B means that B ≥ 0 and therefore −B ≤ f (x) ≤ B. x→0 x→0 lim x2 f (x) = 0. then x x x→0 x→0 1 by the Squeeze Theorem. then −|x| ≤ x sin ≤ |x|. f (x) ≤ x2 + 1 is in fact true for all x. or f (x) ≤ x2 + 1 when f (x) − 1 > 0. 2 x→0 x 1 41. Since |f (x) − 1| ≤ x2 . then by the Squeeze Theorem. Since |x| = . Let t = π − (π/x). lim = lim = 2. we get: 2 x cos(π/x) (sin t)(π − 2t) sin t π − 2t π lim = lim = lim · lim = x→2 x−2 t→0 4t t→0 t t→0 4 4 sin x 49. x>0 sin x 50. x = t + and we have the following substitutions: 4 4 √ √ π π π 2 2 sin x = sin(t + ) = sin t cos + cos t sin = sin t + cos t 4 4 4 √2 √2 π π π 2 2 cos x = cos(t + ) = cos t cos − sin t sin = cos t − sin t 4 !4 4 2 ! 2 √ √ √ √ 2 2 2 2 √ sin x − cos x = sin t + cos t − cos t − sin t = 2 sin t 2 2 2 2 √ sin x − cos x 2 sin t √ With these substitutions. Therefore π/x = π − t and sin(π/x) = sin(π − t) = sin t. x→0+ x x→0 − x x→0 x . we get: x−π t t lim = lim = lim x→π tan 2x t→0 tan(2t + 2π) t→0 tan 2t 1 1 1 1 = lim   = lim = = t→0 tan 2t t→0 1 sin 2t 1·2 2 · t cos 2t t 47. then lim does not exist.4.org 2. Substituting in the same way as in Problem 47. Let t = x − . giving us: π−t sin(π/x) (sin t)(π − t) sin t lim = lim = lim · lim (π − t) = 1 · π = π x→1 x − 1 t→0 t t→0 t t→0 π π 48. Substituting. x<0 x→0 x sin |x| sin x lim+ = lim+ =1 x→0 x x→0 x sin |x| sin(−x) − sin x sin x lim = lim− = lim− = − lim− = −1 x→0− x x→0 x x→0 x x→0 x sin |x| sin |x| sin |x| Since lim 6= lim . knowing that lim = 1 means: −x. Thus. TRIGONOMETRIC LIMITS 91 π π 45. x = t + π. we t can derive x − 1 = . www. f is continuous at x = 0 because lim = 1 = f (0). Thus. Let t = − . In addition. x→0 x  x. x→π/4 x − (π/4) t→0 t 46.elsolucionario. Let t = x − π. −∞ 2. −∞ 5. ∞ 3.92 CHAPTER 2. lim √ = lim = x→−∞ 23x x→−∞ 2 2     3x x−1 3 1 − 1/x 1 5 15. ∞ 6. lim 3 = lim 3 = 3 − =− x→−∞ 7 − 16x x→−∞ 7/x − 16 16 2 √ √

x + x2 + 1 √ −1 19. ∞ 8. lim√3 + √ 5 =0 x→−∞ x x √ √ 8− x (8/ x) − 1 1 13. lim √ = lim √ =− x→∞ 1 + 4 x x→∞ (1/ x) + 4 4 √ √ 1+73x 1/ 3 x + 7 7 14. lim x − + 1 = lim x − x2 +1 · √ x2 = lim √ =0 x→∞ x→∞ x+ x +1 2 x→∞ x + x2 + 1 p  p  √x2 + 5x + x 20. −∞ 7. lim sin = lim sin = sin lim =− x→−∞ 3 − 6x x→−∞ 3/x − 6 x→−∞ 3/x − 6 2 . 5   6 1 12. lim = lim = x→∞ 4x + 5 2 x→∞ 4 + 5/x 2 4 x2 1 10. lim = lim = · 23 = x→∞ 3x + 1 2x2 + x x→∞ 3 + 1/x 2 + 1/x 3 3 r s r √ 3x + 2 3 + 2/x 1 2 17. LIMIT OF A FUNCTION 2. lim cos = cos lim =1 x→∞ x x→∞ x        πx π π 1 22. ∞ 4. lim x + 5x − x = lim 2 x + 5x − x · √ 2 x→∞ x→∞ x2 + 5x + x 5x 5 5 = lim √ = lim p = x→∞ x + 5x + x x→∞ 1 + 5/x + 1 2 2      5 5 21.5 Limits that Involve Infinity 1. lim = lim =∞ x→∞ 1 + x−2 x→∞ 1/x2 + 1/x4 11. −∞ x2 − 3x 1 − 3/x 1 9. lim − = lim − =3− = x→∞ x + 2 2x + 6 x→∞ 1 + 2/x 2 + 6/x 2 2   3   3 4x2 + 1 x 1 4 + 1/x2 1 8 16. lim = lim = = x→∞ 6x − 8 x→∞ 6 − 8/x 2 2 r s r 2x − 1 2 − 1/x 2 1 18. Start with √ = p . From this.2. lim x =   =   x→−∞ e + e−x lim ex + lim e−x 0+ lim e−x x→−∞ x→−∞ x→−∞ −e−x = lim = lim −1 = −1 x→−∞ e−x x→−∞       ex − e−x lim e − lim e−x x lim ex − 0 lim = x→∞  x→∞  = x→∞  x→∞ ex + e−x lim ex + lim e−x lim ex + 0 x→∞ x→∞ x→∞ ex = lim x = lim 1 = 1 x→∞ e x→∞ . 3 x→∞ x→∞ 3 + 1/x2 3 3   6 3 −5 + + −5x2 + 6x + 3 |x| x2 −5 − 6/x + 3/x2 28. Start with √ = p . lim f (x) = lim p = x4 + x2 + 1 1 + 1/x2 + 1/x4 x→−∞ x→−∞ 1 + 1/x2 + 1/x4 −5 −5 + 6/x + 3/x2 −5 √ = −5 and lim f (x) = lim p = √ = −5. lim f (x) = lim p = −√ = 3x + 1 2 3 + 1/x 2 x→−∞ x→−∞ 3 + 1/x 2 3 √ √ 2 3 2 + 1/x 2 2 3 − and lim f (x) = lim p =√ = . LIMITS THAT INVOLVE INFINITY 93         x x    |x|    23. From this. From this. Start with =   . 5 x→∞ x→∞ 5 − 1/x 5 5   2x 1 + 2x + 1 |x| |x| −2 − 1/x 2 27. lim f (x) = lim = = 5x − 1 5x 1 x→−∞ x→−∞ −5 + 1/x −5 − p|x| |x| √ 3 9 + 6/x2 9 3 − and lim f (x) = lim = = .5. Start with √ =p . x→∞ x→∞ 1 + 1/x2 √ p p √ 9x2 + 6 9 + 6/x2 9 + 6/x2 9 26. 1 x→∞ x→∞ 1 + 1/x2 + 1/x4 1       lim e x − lim e −x 0 − lim e −x ex − e−x x→−∞ x→−∞ x→−∞ 29. lim sin−1 √ x = lim sin−1  p  = lim sin−1  p −x  x→−∞ 4x + 1 2 x→−∞  4 + 1/x2  x→−∞  4 + 1/x2  " !#   −1 1 π = sin −1 lim p = sin −1 − =− x→−∞ 4 + 1/x 2 2 6        x 1 1 24. From this. lim f (x) = lim p = −4 and x +1 2 1 + 1/x 2 x→−∞ x→−∞ 1 + 1/x2 4 + 1/x lim f (x) = lim p = 4. lim ln = lim ln = ln lim = ln 1 = 0 x→∞ x+8 x→∞ 1 + 8/x x→∞ 1 + 8/x   4x 1 + 4x + 1 |x| |x| −4 − 1/x 25. lim = lim = lim = −1 x→−∞ x−5 x→−∞ x − 5 x→−∞ 1 − 5/x |x − 5| x−5 lim = lim =1 x→∞ x − 5 x→∞ x − 5 |4x| + |x − 1| −4x − (x − 1) −5x + 1 32. www. -5 5 Vertical asymptote: x = −1 Horizontal asymptote: none -5 . LIMIT OF A FUNCTION   lim 2e−x 2e−x x→−∞ 30.org 94 CHAPTER 2. -5 5 Vertical asymptote: none Horizontal asymptote: y=0 -5 5 35. lim 1+ x =1+     x→−∞ e + e−x lim e + x lim e −x x→−∞ x→−∞ lim 2e −x 2e−x x→−∞ =1+   = 1 + lim = 1 + lim 2 = 3 x→−∞ e−x x→−∞ 0+ lim e−x x→−∞   lim 2e−x 2e−x x→∞ 0 lim 1+ =1+     = 1 + lim x = 1 x→∞ e + e−x x lim ex + lim e−x x→∞ e x→∞ x→∞ |x − 5| −x + 5 −1 + 5/x 31. -5 5 Vertical asymptote: none Horizontal asymptote: y=0 -5 5 34.elsolucionario. lim = lim = lim x→−∞ x x→−∞ x x→−∞ x −5 + 1/x = lim = −5 x→−∞ 1 |4x| + |x − 1| 4x + x − 1 5x − 1 5 − 1/x lim = lim = lim = lim =5 x→∞ x x→∞ x x→∞ x x→∞ 1 5 33. -5 5 Vertical asymptote: x = 0. x = 2 Horizontal asymptote: y=0 -5 5 38. y = 1 -10 .5. -5 5 Vertical asymptote: x = −1 Horizontal asymptote: y=1 -5 5 37. LIMITS THAT INVOLVE INFINITY 95 5 36. Vertical asymptote: x = −1. -5 5 Vertical asymptote: x=1 Horizontal asymptote: y=1 -5 10 40. -5 5 Vertical asymptote: none Horizontal asymptote: y=4 -5 5 39. 10 Vertical asymptote: x=0 Horizontal asymptote: y = −1 -10 10 -10 10 41. y = 1 -10 10 42.2. x = 1 -10 10 Horizontal asymptote: y = −1. Vertical asymptote: none Horizontal asymptote: y = −1. v→c 1 − v 2 /c2 v→c 1 − 1 v→c− 0 . lim x sin = lim x sin = lim x(3/x) x→∞ x x→∞ x 3/x x→∞ 3/x       3 sin 3/x sin 3/x = lim x · lim = lim 3 lim x→∞ x x→∞ 3/x x→∞ x→∞ 3/x At this point. (a) 2 (b) −∞ (c) 0 (d) 2 44.96 CHAPTER 2. resulting in:   sin 3/x  sin t lim 3 lim = 3 lim =3 x→∞ x→∞ 3/x t→0 t m0 m0 m0 52. -5 5 -5 5 -5 5 50. -5 5 -5 5 48. (a) ∞ (b) ∞ (c) 1 (d) 3 45. (a) ∞ (b) −∞ (c) 0 (d) 0 5 47. we substitute t = 3/x. m → ∞. -5      3 3 3/x sin 3/x 51. LIMIT OF A FUNCTION 43. (a) −∞ (b) −3/2 (c) ∞ (d) 0 46. -5 5 -5 5 49. so as v → c− . lim− p = lim− √ = lim . -5 5 -5 (a) lim f (x) = ∞ (b) lim f (x) ≈ 2.99995000  x 1 lim cos =1 x→∞ x 5 55.95114995 0. x→∞ 10 100 1000 10000 f (x) 0.99999999 2. (c) g is a slant asymptote to f .99950012 0. the area of the polygon is:     1 2 π π 1 2π n 2π A(n) = 2n r sin cos = nr 2 sin = r2 sin 2 n n 2 n 2 n (b) A(100) ≈ 3.99986667 1.1416r2 (c) Letting x = 2π/n (while noting that n = 2π/x) and substituting into A(n) above. (a) The area of the right triangle shown in Figure 2. we see that:   sin x lim A(n) = πr2 lim = πr2 n→∞ x→0 x   x2 57. (a) lim [f (x) − g(x)] = lim − (x − 1) x→±∞ x→±∞ x + 1  2  x (x − 1)(x + 1) x2 − (x2 − 1) = lim − = lim x→±∞ x + 1 x+1 x→±∞ x+1 1 = lim =0 x→±∞ x + 1 (b) The graphs of f and g get closer and closer to each other when |x| is large. .18 is r2 sin cos . A(1000) ≈ 3.99501240 0. x→∞ 10 100 1000 10000 f (x) 1.00000000 2.elsolucionario. Since there are 2 n n 2n such right triangles.1395r2 .org 2.5. www. we obtain:   π 2 sin x A(n) = r sin x = πr 2 x x From (10) of Section 2.5.7 (c) lim f (x) = 1 x→−1+ x→0 x→∞ 1 π π 56.4.00000000 2 lim x2 sin =2 x→∞ x2 54. LIMITS THAT INVOLVE INFINITY 97 53. and thus p the horizontal distance between P and Q is |xQ − xP | = | x2P + 1 − xP |. LIMIT OF A FUNCTION 58. we have x2P + 1 = x2Q or xQ = x2P + 1. x2 + 1) while all points Q are of the formp(x. x2 ).98 CHAPTER 2. When the y coordinates of P and Q are the same. All points P are of the form (x. Thus: . !. p . p  √x 2 + 1 + x . . . lim | x + 1 − x| = lim . 2 x +1−x 2 √ . x→∞ x→∞ . x2 + 1 + x . . 2 . . . . x + 1 − x2 . . 1 . . = lim . √ . . = lim . √ . x→∞ x +1+x 2 . = 0. x→∞ x + 1 + x. 7. 3. 5. Choose δ = /2. 4. Choose δ = /6. . |10 − 10| = 0 <  for any choice of δ. |9 − 6x − 3| = |6 − 6x| = 6|x − 1| <  whenever 0 < |x − 1| < /6.6 Limits — A Formal Approach 1. |x − 4 − (−4)| = |x − 0| <  whenever 0 < |x − 0| < . |x − 3| <  whenever 0 < |x − 3| < . 8. Choose δ = /3. Choose δ = . |x + 6 − 5| = |x + 1| <  whenever 0 < |x − (−1)| < . |π − π| = 0 <  for any choice of δ. |3x + 7 − 7| = 3|x − 0| <  whenever 0 < |x − 0| < /3. |2x − 8| = 2|x − 4| <  whenever 0 < |x − 4| < /2. 2. Choose δ = . 6. Choose δ = . 2 2. . . 2x − 3 1 . . 1 1 9. . − . . 4 4 4 2 . Choose δ = 2. = |2x − 4| = |x − 2| <  whenever 0 < |x − 2| < 2. . . . 1 . . |8(2x + 5) − 48| = |16x − 8| = 16 . 10. x − . <  whenever 0 < |x − 2| < /16. 2 . Chose δ = /16. 2 . . x − 25 . 11. . . − (−10). . = |x − 5 + 10| = |x − (−5)| <  whenever 0 < |x − (−5)| < . x+5 . Choose δ = . 2  . . . . x − 7x + 12 1 . . . (x − 3)(x − 4) 1 . 12. . . − = . + . = 1 |x − 4 + 1| = 1 |x − 3| <  whenever 2x − 6 2 . . 2(x − 3) 2. Choose δ = 2. 2 2 0 < |x − 3| < 2. . 5 . . 8x + 12x4 . . 13. . 4 − 12 . = |8x + 12 − 12| = 8|x − 0| <  whenever 0 < |x − 0| < /8. . Choose δ = /8. x . 3 . . . . 2x + 5x2 − 2x − 5 . . (2x + 5)(x2 − 1) . 14. . . − 7 . =. . . − 7. = |2x + 5 − 7| = |2x − 2| = 2|x − 1| <  . √ √ 16. √ √ 15. x −1 2 x −1 2 whenever 0 < |x − 1| < /2. Choose δ = . |x2 − 0| = |x − 0|2 <  whenever 0 < |x − 0| < . |8x3 − 0| = 8|x − 0|3 <  whenever 0 < |x − 0| < 3 /2. . Choose δ = 3 /2. Choose δ = /2. We need √ to show | x − a| <  whenever 0 < |x − a| < δ for an appropriate choice of δ. x→a 26. /10}. Choose δ = min{1. | 5x − 0| = 5|x − 0|1/2 <  whenever 0 < x < 2 /5. |2x − 1 − (−1)| = |2x| = 2|x − 0| <  whenever 0 − /2 < x < 0. so |x + 3| < 7. |x2 − 9| = |x − 3||x + 3| < 7|x − 3| <  whenever |x − 3| < /7. Note that |x2 +2x−35| = |x−5||x+7| and consider only values of x for which |x−5| < 1. Thus. Choose δ = /2. lim x = a.6. Then. Without loss of generality. whenever 0 < |x − 2| < δ. Choose δ = . 1/3 < 1/x < 1. Thus |2x2 + 4 − 12| = 2|x − 2||x + 2| < 10|x − 2| <  whenever |x − 2| < /10. for δ = 2. Then |x − 2| < 1 or 1 < x < 3. 20. √ √ 18. for an appropriate choice of δ. we may assume that δ < 1. 22. Note that |x2 − 2x + 4 − 3| = |x − 1|2 <  whenever |x − 1| < . For δ = a. Choose δ = min{1. We need to show that |1/x − 1/2| < . so |x + 2| < 5. Note that |x2 − 9| = |x − 3||x + 3| and consider only values of x for which |x − 3| < 1. Thus. Note that |2x2 + 4 − 12| = 2|x2 − 4| = 2|x − 2||x + 2| and consider only values of x for which |x − 2| < 1. /7}. we have . 24. Then 1 < x < 3 and 3 < x + 2 < 5. 19. √ √ 25. /13}. |3 − 3| = 0 <  whenever x > 1. 21. so |x + 7| < 13. Choose δ = 2 /2. Choose δ = min{1. | 2x − 1 − 0| = 2|x − 1/2|1/2 <  whenever 1/2 < x < 1/2 + 2 /2. Then 2 < x < 4 and 5 < x + 3 < 7. LIMITS — A FORMAL APPROACH 99 √ √ 17. Choose δ = 2 /5. Thus |x2 + 2x − 35| = |x − 5||x + 7| < 13|x − 5| whenever |x − 5| < /13. Then 4 < x < 6 and 11 < x + 7 < 13. For these values of x.2. for any choice of δ. we have √ √ √ √ √ √ √ x+ a |x − a| |x − a| a | x − a| = | x − a| · √ √ =√ √ < √ < √ = x+ a x+ a a a √ √ whenever 0 < |x − a| < δ. √ √ 23. .   . 1 1. 1 1 1 1 . − . = . x 2 . Then there exists δ > 0 such that |f (x) − L| < 1 whenever x→1 0 < |x − 1| < δ. Take  = 1. Assume lim f (x) = L. To the right of 1. Since 0 < |1 + δ/2 − 1| = |δ/2| < δ. 2 x |2 − x| < 2 (1)|x − 2| < 2 (2) =  whenever 0 < |x − 2| < δ. choose x = 1 + δ/2. we must have |f (1 − δ/2) − L| = |2 − L| < 1. Thus. choose x = 1 − δ/2. or 1 < L < 3. lim 1/x = 1/2. x→2 27. Since 0 < |1 − δ/2 − 1| = | − δ/2| < δ. we must have |f (1 + δ/2) − L| = |0 − L| = |L| < 1. . or −1 < L < 1. To the left of 1. or 0 < L < 2. x→3 29. To the left of 3. we must have |f (3 − δ/2) − L| = |1 − L| = |L − 1| < 1. choose x = δ/2. Since 0 < |δ/2 − 0| = |δ/2| < δ. or 2/δ − 1 < L < δ/2 + 1.elsolucionario. Then there exists δ > 0 such that |f (x) − L| < 1 whenever x→0 0 < |x − 0| < δ. To the right of 3. Then there exists δ > 0 such that |f (x) − L| < 1 whenever x→0 0 < |x − 0| < δ. Assume lim f (x) = L. Take  = 1. Assume lim f (x) = L. Take  = 1. choose x = δ/2. or −2/δ − 1 < L < −2/δ + 1. x→0 30. LIMIT OF A FUNCTION Since no L can satisfy the conditions that −1 < L < 1 and 1 < L < 3. Since 0 < |δ/2 − 0| = |δ/2| < δ. Since 0 < |3 − δ/2 − 3| = | − δ/2| < δ. we may assume that δ < 2. Assume lim f (x) = L. Since |f (x) − L| < 1 for all x such that 0 < |x| < δ. or −1 − δ/2 < L < 1 − δ/2. we conclude that lim f (x) does not exist. Since no L can satisfy the conditions that 1 − δ/2 < L < 3 − δ/2 and −1 − δ/2 < L < 1 − δ/2. we must have |f (δ/2) − L| = |2/δ − L| = |L − 2δ| < 1. choose x = −δ/2. Since 0 < | − δ/2 − 0| = | − δ/2| < δ. or 1 − δ/2 < L < 3 − δ/2. Take  = 1. . choose x = 3 − δ/2. www. To the right of 0. Then there exists δ > 0 such that |f (x) − L| < 1 whenever x→3 0 < |x − 3| < δ. Since 0 < |3 + δ/2 − 3| = |δ/2| < δ. To the right of 0. choose x = −δ/2. we must have |f (−δ/2) − L| = | − 2/δ − L| = |L + 2/δ| < 1. we conclude that lim f (x) does not exist. Since 0 < | − δ/2 − 0| = |δ/2| < δ. To the left of 0. choose x = 3 + δ/2. we must have |f (3 + δ/2) − L| = | − 1 − L| = |L + 1| < 1.org 100 CHAPTER 2. To the left of 0. we conclude that lim f (x) does not exist. or −2 < L < 0. x→1 28. we must have |f (−δ/2) − L| = | − δ/2 − L| < 1. Since no L can satisfy the conditions that −2 < L < 0 and 0 < L < 2. we must have |f (δ/2) − L| = |2 − δ/2 − L| < 1. x→0 31.5(i).6. for any  > 0 we must find an N > 0 such that . these imply 0 < L < 2δ + 1 and −2/δ − 1 < L < 0. By Definition 2. lim f (x) does not exist. Since it is impossible for L to satisfy both of these inequalities. we have 1 < 2/δ or 0 < 2/δ − 1 and −1 > −2/δ or 0 > −2/δ + 1. Having established 2/δ − 1 < L < δ/2 + 1 and −2/δ − 1 < L < −2/δ + 1. LIMITS — A FORMAL APPROACH 101 Since we assumed δ < 2.2.6. . . 5x − 1 5 . . . <  whenever x > N. . 2x + 1 − 2. Now by considering x > 0. . . . . . 5x − 1 5 . . −7 . 7 7 . . . . . 2x + 1 − 2 . = . 4x + 2 . = 4x + 2 < 4x <  whenever x > 7/4.6. for any  > 0 we must find an N > 0 such that .5(i). By Definition 2. 32. choose N = 7/4. Hence. . . 2x 2 . . . . <  whenever x < N. 3x + 8 − 3. . Now by considering x > 0. . . . . 2x 2 . . . . −16 . . 16 16 . . 3x + 8 − 3 . = . 9x + 24 . choose N = 16/9. By Definition 2. Hence.5(ii). 33. = 9x + 24 < 9x <  whenever x > 16/9.6. for any  > 0 we must find an N < 0 such that . . . 10x . . − 10 . <  whenever x < N. . x − 3 . Now by considering x < 0. . . . . . . . 10x . . 30 . . 30 . 30 30 . . . . . . . x − 3 − 10. = . x − 3 . = . −(−x + 3) . choose N = −30/.5(ii).6. By Definition 2. 34. Hence. for any  > 0 we must find an N < 0 such that . = −x + 3 < x <  whenever x < −30/. . . x2 . . . . x2 + 3 − 1. Now by considering x < 0. . <  whenever x < N. . . . . x2 . . . −3 . . =. . 3 3 . x2 + 3 − 1. . x2 + 3 . Hence. . choose N = − 3/. = x2 + 3 < x2 <  p p whenever x2 > 3/ or x < − 3/. We need to show |f (x) − 0| = |f (x)| <  whenever 0 < |x − 0| = |x| < δ for an appropriate choice of δ. -10 -5 5 10 change in x = h = 2.7 The Tangent Line Problem 10 1. -10 -5 5 10 change in x = h = −1 − (−2) = 1 change in y = f (−2 + 1) − f (−2) = −1 − (−8) = 7 change in y 7 -10 msec = = =7 change in x 1 5 4.75 − 5 = −2. For δ = .1 change in y = f (1 + 0. ( |x|.5) − f (2) = 2.25 change in y −2. -5 5 change in x = h = 1 − 0.5 − 2 = 0.5 change in y = f (2 + 0. -10 -5 5 10 change in x = h = 0 − (−1/4) = 1/4 change in y = f (0 + 1/4) − f (0) = 17/16 − 0 = 17/16 change in y 17/16 17 -10 msec = = = change in x 1/4 4 10 3.5 10 2. x irrational Thus.5 change in x 0. 0.1) − f (1) = 10/11 − 1 = −1/11 -5 change in y −1/11 10 msec = = =− change in x 1/10 11 .102 CHAPTER 2.25 -10 msec = = = −4. lim f (x) = 0. LIMIT OF A FUNCTION 35. x→0 2. x rational |f (x)| = <  whenever 0 < |x| < δ.9 = 0. −17).7. f (a + h) = f (−1 + h) = −3(h − 1)2 + 10 f (a + h) − f (a) = [−3(h − 1)2 + 10] − 7 = [(−3h2 + 6h − 3) + 10] − 7 = −3h2 + 6h = h(6 − 3h) f (a + h) − f (a) h(6 − 3h) mtan = lim = lim = lim (6 − 3h) = 6 h h→0 h→0 h h→0 With point of tangency (−1. 10. f (a) = f (1) = −2. we have y − 7 = 6(x + 1) or y = 6x + 13.org 2. www. f (a) = f (−1) = 7.elsolucionario. 3). change in x = h = − − − = –π –π 3 2 6 3  π π  π  π 1 change in y = f − + −f − = cos − − √ 3 6√ 3 6 2 -3 3 1 3−1 = − = 2 2 2√ √ change in y ( 3 − 1)/2 3 3−3 msec = = = change in x π/6 π 7. we have y + 17 = 9(x + 2) or y = 9x + 1. f (a + h) = f (1 + h) = (h + 1)2 − 3(h + 1) f (a + h) − f (a) = [(h + 1)2 − 3(h + 1)] − (−2) = (h2 − h − 2) − (−2) = h2 − h = h(h − 1) f (a + h) − f (a) h(h − 1) mtan = lim = lim = lim (h − 1) = −1 h h→0 h→0 h h→0 With point of tangency (1. −2). f (a + h) = f (−2 + h) = −(h − 2)2 + 5(h − 2) − 3 f (a + h) − f (a) = [−(h − 2)2 + 5(h − 2) − 3] − (−17) = (−h2 + 9h − 17) − (−17) = −h2 + 9h = h(9 − h) f (a + h) − f (a) h(9 − h) mtan = lim = lim = lim (9 − h) = 9 h→0 h h→0 h h→0 With point of tangency (−2. THE TANGENT LINE PROBLEM 103 3 2π π π 5. we have y + 2 = −(x − 1) or y = −x − 1. 8. we have y − 3 = 6(x − 3) or y = 6x − 15. f (a) = f (3) = 3. . f (a + h) = f (3 + h) = (h + 3)2 − 6 f (a + h) − f (a) = [(h + 3)2 − 6] − 3 = [(h2 + 6h + 9) − 6] − 3 = h2 + 6h = h(h + 6) f (a + h) − f (a) h(h + 6) mtan = lim = lim = lim (h + 6) = 6 h→0 h h→0 h h→0 With point of tangency (3. 7). change in x = h = − = π π 3 2 6 2 π π π 2 √ change in y = f + −f = sin π − 1 = 3/2 − 1 -3 2 6 √2 3 √ change in y 3/2 − 1 3 3−6 msec = = = change in x π/6 π 3 π  π π 6. 9. f (a) = f (−2) = −17. f (a) = f (1/2) = −3. we have y − 4 = −4(x − 2) or y = −4x + 12. f (a) = f (2) = 4. f (a + h) = f (2 + h) = −2(h + 2)3 + (h + 2) f (a + h) − f (a) = [−2(h + 2)3 + (h + 2)] − (−14) = (−2h3 − 12h2 − 23h − 14) − (−14) = h(−2h2 − 12h − 23) f (a + h) − f (a) h(−2h2 − 12h − 23) mtan = lim = lim h→0 h h→0 h = lim (−2h2 − 12h − 23) = −23 h→0 With point of tangency (2. 12. . LIMIT OF A FUNCTION 11.104 CHAPTER 2. 4). we have y − 1 = 2(x − 0) or y = 2x + 1. we have y + 3 = 6(x − 1/2) or y = 6x − 6. −1/2). 1 15. f (a + h) = f (h) = (h − 1)2 1 −h2 + 2h h(2 − h) f (a + h) − f (a) = − 1 = = (h − 1)2 (h − 1)2 (h − 1)2   f (a + h) − f (a) h(2 − h) 1 2−h mtan = lim = lim · = lim =2 h→0 h h→0 (h − 1)2 h h→0 (h − 1)2 With point of tangency (0. 2 2 2 4 14. f (a) = f (2) = −14. 1). f (a) = f (0) = 1. −3). we have y + = − (x + 1) or y = − . f (a + h) = f (−1 + h) = 2(h − 1)   1 1 1+h−1 h f (a + h) − f (a) = − − = = 2(h − 1) 2 2(h − 1) 2(h − 1)   f (a + h) − f (a) h 1 mtan = lim = lim · h→0 h h→0 2(h − 1) h 1 1 = lim =− h→0 2(h − 1) 2 1 1 x With point of tangency (−1. 1 13. we have y + 14 = −23(x − 2) or y = −23x + 32. f (a + h) = f (1/2 + h) = 8(h + 1/2)3 − 4 f (a + h) − f (a) = [8(h + 1/2)3 − 4] − (−3) = (8h3 + 12h2 + 6h − 3) − (−3) = 2h(4h2 + 6h + 3) f (a + h) − f (a) 2h(4h2 + 6h + 3) mtan = lim = lim h→0 h h→0 h = lim 2(4h + 6h + 3) = 6 2 h→0 With point of tangency (1/2. f (a) = f (−1) = −1/2. f (a + h) = f (2 + h) = (h + 2) − 1 4 4 − 4h − 4 −4h f (a + h) − f (a) = −4= = (h + 2) − 1 h+1 h+1   f (a + h) − f (a) −4h 1 −4 mtan = lim = lim · = lim = −4 h→0 h h→0 h + 1 h h→0 h + 1 With point of tangency (2. −14). we have y − 2 = (x − 4) or y = x + 1. 2 2 6 2 12 2 . f (a + h) = f (π/6 + h) = sin(π/6 + h) π  1 π π 1 f (a + h) − f (a) = sin + h − = sin cos h + cos sin h − 6 √ 2 6 6 √ 2 1 3 1 1 3 = cos h + sin h − = (cos h − 1) + sin h 2 2 2 2 2 ! √ f (a + h) − f (a) 1 cos h − 1 3 sin h mtan = lim = lim · + · h→0 h h→0 2 h 2 h √ √ = (1/2)(0) + ( 3/2)(1) = 3/2 √  √ √ 1 3 π 3 3π 1 With point of tangency (π/6. 2 2 2 19. 1/2). f (a + h) = f (1 + h) = √ h+1 √ √ √ 1 1− h+1 1− h+1 1+ h+1 f (a + h) − f (a) = √ −1= √ = √ · √ h+1 h+1 h+1 1+ h+1 1−h−1 −h =√ =√ h+1+h+1 h+1+h+1   f (a + h) − f (a) −h 1 mtan = lim = lim √ · h→0 h h→0 h+1+h+1 h −1 1 = lim √ =− h→0 h+1+h+1 2 1 1 3 With point of tangency (1. f (a + h) = f (−1 + h) = 4 − −1 + h   8 −8 −8 − 8h + 8 −8h f (a + h) − f (a) = 4 − − 12 = −8= = −1 + h −1 + h h−1 h−1   f (a + h) − f (a) −8h 1 −8 mtan = lim = lim · = lim =8 h→0 h h→0 h − 1 h h→0 h − 1 With point of tangency (−1.7. THE TANGENT LINE PROBLEM 105 8 16. we have y − 12 = 8(x + 1) or y = 8x + 20. we have y − = x− or y = x− + . we have y − 1 = − (x − 1) or y = − x + . √ 17. f (a) = f (1) = 1. f (a) = f (π/6) = 1/2. f (a) = f (−1) = 12. 12). 2). f (a + h) = f (4 + h) = 4 + h √ √ √ 4+h+2 4+h−4 h f (a + h) − f (a) = 4 + h − 2 = ( 4 + h − 2) √ =√ =√ 4+h+2 4+h+2 4+h+2   f (a + h) − f (a) h 1 mtan = lim = lim √ · h→0 h h→0 4+h+2 h 1 1 = lim √ = h→0 4+h+2 4 1 1 With point of tangency (4. f (a) = f (4) = 2.2. 1). 4 4 1 18. so its equation is 0−4 y−0= (x − 2) or y =x−2 2−6 The line’s y-intercept is (0. 0) and (6. −2). 6) is m = (6 − 1)/(4 − 1) = 5/3. f (a) = a2 . we first find the slope of the tangent line at the “general point” (a. The slope of the tangent at the blue point (3. 1) and (1. 24. then this line is tangent to the graph. LIMIT OF A FUNCTION √ 20. Since there is more than one line. f (a)). f (a + h) = f (π/4 + h) = cos(π/4 + h) π  √2 π π √ 2 f (a + h) − f (a) = cos +h − = cos cos h − sin sin h − √ 4 √ 2 √ 4 √ 4 2 2 2 2 2 = cos h − sin h − = (cos h − sin h − 1) 2 2 2 √ 2   f (a + h) − f (a) 2 cos h − 1 sin h mtan = lim = lim − h→0 h h→0 2 h h √ √ = ( 2/2)(0 − 1) = − 2/2 √ √  √ √ √ 2 2 π 2 2π With point of tangency (π/4. Since the slopes are not equal. 1) and (4. 23. 2 21. 22. 9) and (1. The slope of the line through (−1. then this line is tangent to the graph. 0) are on the tangent line. f (a) = f (π/4) = 2/2. f (a) = f (1) = 1. f (a + h) = (h + a)2 f (a + h) − f (a) = [(h + a)2 ] − (a2 ) = (h2 + 2ha + a2 ) − a2 = h(h + 2a) f (a + h) − f (a) h(h + 2a) mtan = lim = lim = lim (h + 2a) = 2a h→0 h h→0 h h→0 Now that we have determined that mtan = 2a. The slope of the line through (1. f (a + h) = f (1 + h) = (h + 1)2 f (a + h) − f (a) = [(h + 1)2 ] − 1 = (h2 + 2h + 1) − 1 = h(h + 2) f (a + h) − f (a) h(h + 2) mtan = lim = lim = lim (h + 2) = 2 h→0 h h→0 h h→0 The slope of the tangent at the blue point (1. then the slope of the tangent at the blue point (−1. −3) is m = (9 + 3)/(3 − 1) = 6. 4) are on the tangent line.elsolucionario. 1) is 2. we have y − =− x− or y = − x+ + √ 2 2 4 2 8 2 . 2/2). 1) is mtan (−1) = 2(−1) = −2. The slope of the line through (3. We know that the points (0. then this line is not tangent to the graph.org 106 CHAPTER 2. We know that the points (2. Since the slopes are equal. 4) and (7. Since the slopes are equal. www. so its equation is 4−0 4 y−0= (s − 7) or y =− x+4 0−7 7 . 9) is mtan (3) = 2(3) = 6. −3) is m = (−3 − 1)/(1 + 1) = −2. −94). then 4 48 f (−5) = − (−5) + 4 = 7 7 25. 27. 26. the tangent line is horizontal at (3. f (a) = a3 − 3a. yielding 3a2 = 3 and a = ±1. f (−5)) is on this tangent line. the tangent line is horizontal at (−1. so we substitute and solve mtan = 0 = 3a2 − 3. . f (a + h) = −(h + a)2 + 6(h + a) + 1 f (a + h) − f (a) = [−(h + a)2 + 6(h + a) + 1] − (−a2 + 6a + 1) = −h2 − 2ha − a2 + 6h + 6a + 1 − (−a2 ) − 6a − 1 = −h2 − 2ha + 6h = h(−h − 2a + 6) f (a + h) − f (a) h(−h − 2a + 6) mtan = lim = lim h→0 h h→0 h = lim (−h − 2a + 6) = −2a + 6 h→0 The tangent line is horizontal when mtan = 0. yielding 2a = 6 and a = 3. Thus. f (a + h) = 2(h + a)2 + 24(h + a) − 22 f (a + h) − f (a) = [2(h + a)2 + 24(h + a) − 22] − (2a2 + 24a − 22) = 2h2 + 4ha + 2a2 + 24h + 24a − 22 − 2a2 − 24a − (−22) = 2h2 + 4ha + 24h = h(2h + 4a + 24) f (a + h) − f (a) h(2h + 4a + 24) mtan = lim = lim h→0 h h→0 h = lim (2h + 4a + 24) = 4a + 24 h→0 The tangent line is horizontal when mtan = 0. f (a) = 2a2 + 24a − 22. f (a + h) = (h + a)3 − 3(h + a) f (a + h) − f (a) = [(h + a)3 − 3(h + a)] − (a3 − 3a) = h3 + 3h2 a + 3ha2 + a3 − 3h − 3a − a3 − (−3a) = h3 + 3h2 a + 3ha2 − 3h = h(h2 + 3ah + 3a2 − 3) f (a + h) − f (a) h(h2 + 3ah + 3a2 − 3) mtan = lim = lim h→0 h h→0 h = lim (h2 + 3ah + 3a2 − 3) = 3a2 − 3 h→0 The tangent line is horizontal when mtan = 0. so we substitute and solve mtan = 0 = 4a + 24. yielding 4a = −24 and a = −6.2. the tangent line is horizontal at (−6. 2) and (1. Thus. so we substitute and solve mtan = 0 = −2a + 6.7. −2). Thus. THE TANGENT LINE PROBLEM 107 Since the point of tangency (−5. 10). f (1)) = (1. f (−6)) = (−6. f (a) = −a2 + 6a + 1. f (3)) = (3. f (−1)) = (−1. f (0)) = (0.9∆t2 − 4.9∆t2 − 4.275 = −4.108 CHAPTER 2. 920 km/h = . f (a + h) = −(h + a)3 + (h + a)2 f (a + h) − f (a) = [−(h + a)3 + (h + a)2 ] − (−a3 + a2 ) = −h3 − 3h2 a − 3ha2 − a3 + h2 + 2ah + a2 − (−a3 ) − a2 = −h3 − 3h2 a − 3ha2 + h2 + 2ah = h(−h2 − 3ah − 3a2 + h + 2a) f (a + h) − f (a) h(−h2 − 3ah − 3a2 + h + 2a) mtan = lim = lim h→0 h h→0 h = lim (−h − 3ah − 3a + h + 2a) = −3a + 2a 2 2 2 h→0 The tangent line is horizontal when mtan = 0. change of distance 3500 km 31. f (2/3)) = (2/3. (a) ∆s = s(t0 + ∆t) − s(t0 ) = f (1/2 + ∆t) − f (1/2) = −4.9) = −4. LIMIT OF A FUNCTION 28.9∆t − 4. f (a) = −a3 + a2 .9∆t v(1/2) = lim = lim = lim (−4. vave = = 1 = = = 6 mi/h change in time 3 6 h − 1 12 h 19/6 h − 3/2 h 5/3 h 33. ∆t→0 ∆t ∆t→0 (5∆t + 1)∆t ∆t→0 5∆t + 1 35. t ≈ 3.9(1/2 + ∆t)2 + 122.9∆t The instantaneous velocity at t = 1/2 is ∆s −4. vave = = = = = 45 mi/h change in time 40 s (40 s)/(3600 s/h) 1/90 h The car will not be stopped for speeding. vave = = = 58 mi/h change in time 5h change of distance 1/2 mi (1/2 mi) 1/2 mi 30. 0) and (2/3.8 h = 3 h 48 min change in time t change of distance 20 mi − 10 mi 20 mi − 10 mi 10 mi 32. ∆s = s(t0 + ∆t) − s(t0 ) = f (∆t) − f (0) = ∆t2 + −1= = −5 5∆t + 1 5∆t + 1 The instantaneous velocity at t = 0 is ∆s 5∆t3 + ∆t2 − 5∆t 5∆t2 + ∆t − 5 v(0) = lim = lim = lim = −5.9 m/s. 2/3. ∆t→0 ∆t ∆t→0 ∆t ∆t→0 . change of distance 290 mi 29.5 − 121. ∆t→0 ∆t ∆t→0 ∆t ∆t→0 1 5∆t3 + ∆t2 − 5∆t 34. Thus. the tangent line is horizontal at (0. yielding a(3a − 2) = 0 and a = 0. ∆s = s(t0 +∆t)−s(t0 ) = f (3+∆t)−f (3) = [−4(3+∆t)2 +10(3+∆t)+6]−0 = −14∆t−4∆t2 The instantaneous velocity at t = 3 is ∆s −14∆t − 4∆t2 v(3) = lim = lim = lim (−14 − 4∆t) = −14. so we substitute and solve mtan = 0 = −3a2 +2a. 4/27). vave = . org 2.5/4. THE TANGENT LINE PROBLEM 109 (b) The ball hits the ground when s(t) = 0: −4.9(5)2 + 122. t = 5 s.165 ft/s.03) = −33.5 ≈ 6. 37. p (b) Earth: timpact = 2(100)/32 = 2. we obtain t = 2h/g. (a) s(t) = −16t2 + 256t s(2) = −16(22 ) + 256(2) = 448 ft s(6) = −16(62 ) + 256(6) = 960 ft s(9) = −16(92 ) + 256(9) = 1008 ft s(10) = −16(102 ) + 256(10) = 960 ft (b) s(5) = −16(52 ) + 256(5) = 880 ft s(2) = 448 ft [from (a)] change of distance 880 ft − 448 ft vave = = = 144 ft/s change in time 5 s−2 s . ∆s = s(t0 + ∆t) − s(t0 ) = f (5 + ∆t) − f (5) = [−4.9t2 + 122. t2 = 122.08 s p Moon: timpact = 2(100)/5.elsolucionario.03 s 1 1 1 (c) ∆s = s(t0 + ∆t) − s(t0 ) = − g(t0 + ∆t)2 + h − (− gt20 + h) = − g∆t2 − gt0 ∆t 2 2 2 The instantaneous velocity at timpact is 1   ∆s − g∆t2 − gt0 ∆t 1 v(timpact ) = lim = lim 2 = lim − g∆t − gt0 = −gt0 . ∆t→0 ∆t ∆t→0 ∆t ∆t→0 p 36. (a) Setting − 12 gt2 + h = 0 and solving for t > 0.7. (c) Since the ball impacts at t = 5. ∆t→0 ∆t ∆t→0 ∆t ∆t→0 2 (d) The impact velocities are vEarth = −(32)(2.96 ft/s vMoon ≈ −(5. www.5) = −80 ft/s vMars ≈ −(12)(4.5 = 0.5] = −49∆t2 − 49∆t The impact velocity at t = 5 is ∆s −49∆t2 − 49∆t v(5) = lim = lim = lim (−49∆t − 49) = −49 m/s.9.5 s p Mars: timpact = 2(100)/12 ≈ 4.5] − [−4.9(5 + ∆t)2 + 122.08) = −48.5)(6. v0 ≈ 1 ft/s. the height is once again 1008 ft. (a) s(4) ≈ 1. v(t) = 256 − 32t so v(16) = 256 − 32(16) = −256 ft/s. in this case.7 ft s(6) − s(4) 2. the net distance is zero. (f) The velocity is increasing where the slopes of the tangent lines are increasing. Since distance upward is positive and distance downward is negative. the projectile impacts at t = 16 s.3 ft. Since f is even. for 3 < t < 7. After it reaches a maximum height. From (e). we have s(8) = −16(82 ) + 256(8) = 1024 ft. Since s(t) = −16t2 + 256t.7 − 1. the projectile is at a height of 1008 ft on its way upward. t = 256/16 = 16 s (e) For some general time t: ∆s = s(t + ∆t) − s(t) = [−16(t + ∆t)2 + 256(t + ∆t)] − (−16t2 + 256t) = −16∆t2 + 256∆t − 32t∆t = ∆t(−16∆t + 256 − 32t) The instantaneous velocity at a general time t is ∆s ∆t(−16∆t + 256 − 32t) v(t) = lim = lim ∆t→0 ∆t ∆t→0 ∆t = lim (−16∆t + 256 − 32t) = (256 − 32t) ft/s. The slopes m of a tangent line at (a. 38. 39. f (a)) and m0 of a tangent line at (−a. for 0 < t < 3. In this case. s(6) ≈ 2. ∆t→0 (f) From (d).3 (b) vave ≈ = = 0.2.110 CHAPTER 2. at t = 9 s. 16t2 = 256t. (d) t ≈ 3 s (e) The velocity is decreasing where the slopes of the tangent lines are decreasing. (d) The projectile hits the ground when s(t) = 0: −16t2 + 256t = 0.7 ft/s 6−4 2 (c) The instantaneous velocity at t = 0 is the slope of the tangent line to the graph at t = 0. m0 = lim h→0 h h0 →0 h0 As defined in Section 1. an even function is a function which is symmetric with respect to the y-axis: f (−x) = f (x) for all x. in this case. resulting in: f (a − h0 ) − f (a) f (a + [−h0 ]) − f (a) m0 = lim = lim 0 h →0 h0 h0 →0 h0 . LIMIT OF A FUNCTION (c) s(7) = −16(72 ) + 256(7) = 1008 ft s(9) = 1008 ft [from (a)] change of distance 1008 ft − 1008 ft 0 vave = = = = 0 ft/s change in time 9 s−7 s 2 At t = 7 s. (g) The maximum height is reached when v(t) = 0: 256 − 32t = 0 gives us t = 8 s. then f (−a) = f (a) and f (−a + h0 ) = f (−[−a + h0 ]) = f (a − h0 ). it begins to fall downward and. f (−a)) are: f (a + h) − f (a) f (−a + h0 ) − f (−a) m = lim . we apply the substitution h0 = −h to obtain: −[f (a − h0 ) + f (a)] −[f (a + h) − f (a)] m0 = lim 0 = lim =m 0 h →0 h h→0 −h 41. f (−a)) are: f (a + h) − f (a) f (−a + h0 ) − f (−a) m = lim . Since f is odd. an odd function is a function which is symmetric with respect to the origin: f (−x) = −f (x) for all x. and that therefore f has no tangent line at (0. m0 = lim h→0 h 0 h →0 h0 As defined in Section 1.2. The slopes m of a tangent line at (a. resulting in: −f (a − h0 ) − [−f (a)] −[f (a − h0 ) + f (a)] m0 = lim 0 = − lim 0 h →0 h 0 h →0 h0 Without loss of generality. we conclude that lim = h→0 h h2 + |h| lim does not exist. lim x − 5 = 0. False. False. True √ 2. f (a)) and m0 of a tangent line at (−a. we see that h2 + |h| h2 + h lim+ = =h+1=1 h→0 h h whereas h2 + |h| h2 − h lim− = = h − 1 = −1 h→0 h h f (0 + h) − f (0) Since the right-hand and left-hand limits are not equal. then f (−a) = −f (a) and f (−a + h0 ) = −f (−[−a + h0 ]) = −f (a − h0 ). lim− = −1. h→0 h Chapter 2 in Review A. True/False 1. x→0 x . 0). we apply the substitution h0 = −h to obtain: f (a + [−h0 ]) − f (a) f (a + h) − f (a) m0 = lim = lim = −m 0 h →0 h0 h→0 −h 40. To show that the graph of f (x) = x2 +|x| does not possess a tangent line at (0. we examine f (0 + h) − f (0) [(0 + h)2 + |0 + h|] − 0 h2 + |h| lim = lim = lim h→0 h h→0 h h→0 h From the definition of absolute value.CHAPTER 2 IN REVIEW 111 Without loss of generality. 0). x→5+ |x| 3. consider f (x) = . 2. True 13. True ( −1. False. False. . lim f (x) = 4 = f (5). f (3)) is 1. x<0 16. consider f (x) = 1 and g(x) = x − 2. x+1 22. www. 14.org 112 CHAPTER 2. x2 x 1 1 10. x≤3 17. let f (x) = 0. consider f (x) = −x. LIMIT OF A FUNCTION 2 4. since lim [(x − a)f (x)] = [ lim (x − a)][ lim f (x)] = 0 · f (a) = 0. consider f (x) = and a = 0. False. True 8. True 7. and a = 0. True 20. False. g(x) = . x→∞   1 π 5. x 2 tan2 x 11. and a = 0. There is not enough information to determine the value of f (3). False. False. False. since is undefined for x < 0. 12. 1 1 9. x→5 √ x 21. False. 1. the slope m of the tangent line at (3. since f (−1) < 0 and f (1) > 0. x>0 ( 1. lim e2x−x = 0. False. True. False. x>3 18. 15. consider f (x) = . lim tan −1 = .elsolucionario. g(x) = 4 . x→0+ x 2 6. x→a x→a x→a 19. consider f (x) = . True. False. False. 3− 12. 0+ 15. 1/4 11. 0 9.CHAPTER 2 IN REVIEW 113 B. ∞ 8. −2   x2 f (x) x2 16. 1 3. 2 21. by the Squeeze 3 x x→0 3 x→0 f (x) Theorem we have lim = 1. lim f (g(x)) = f ( lim g(x)) = f (−9) = (−9)2 = 81. 4 2. -1/5 4. Fill in the Blanks 1. x→−5 x→−5 . 9 22. 8 19. 1 10. Since f (x) = x2 is continuous. continuous 20. 3/5 7. Since lim 1 − = 1 = lim 1. 0 6. 4 13. -1/2 5. x→0 x2 17. −∞ 14. 10 18. Dividing by x2 we have 1 − ≤ 2 ≤ 1. (i) 9. (e). (d). Exercises 5 5 1. -5 5 -5 5 -5 -5 3 5 3. ∞) . (h) 8. -3 3 -5 -3 5. (g). (e). (j) 5 11. 3 3 6 13. (c). (0. (h) 7. (−1. (b). (a). (f). (d). 2. (c). (b). 4. -5 5 -5 6 12. LIMIT OF A FUNCTION C. (e). (f). The function is continuous everywhere. (b). (h) 6. The function is discontinuous at x = 2 and x = 4. (f) 10. 0). (−∞. 1). −1).114 CHAPTER 2. (e). (c). (a). and (1. x→3+ Thus. f (a + h) = f (1/2 + h) = 2(h + 1/2)2 −1 −1 f (a + h) − f (a) = − (−2) = 2 − (−2) 2(h + 1/2) 2 2h + 2h + 1/2 −1 + 4h2 + 4h + 1 4h(h + 1) = = 2 2h + 2h + 1/2 2 2h + 2h + 1/2 f (a + h) − f (a) 4h(h + 1) mtan = lim = lim =8 h→0 h h→0 (2h2 + 2h + 1/2)h . 17. f (a + h) = f (2 + h) = −3(h + 2)2 + 16(h + 2) + 12 f (a + h) − f (a) = [−3(h + 2)2 + 16(h + 2) + 12] − 32 = −3h2 − 12h − 12 + 16h + 32 + 12 − 32 = −3h2 + 4h = h(−3h + 4) f (a + h) − f (a) h(−3h + 4) mtan = lim = lim =4 h→0 h h→0 h With point of tangency (2. (−∞. −1 21. Therefore: x  + 1. 2. f (a) = f (1/2) = −2. f (a) = f (2) = 32. f (a + h) = f (−1 + h) = (h − 1)3 − (h − 1)2 f (a + h) − f (a) = [(h − 1)3 − (h − 1)2 ] − (−2) = h3 − 3h2 + 3h − 1 − h2 + 2h − 1 − (−2) = h3 − 4h2 + 5h = h(h2 − 4h + 5) f (a + h) − f (a) h(h2 − 4h + 5) mtan = lim = lim =5 h→0 h h→0 h With point of tangency (−1.org CHAPTER 2 IN REVIEW 115 14. . we have y + 2 = 5(x + 1) or y = 5x + 3. f (a) = f (−1) = −2. . For f (x) to be continuous everywhere. www. Solving for a x→3 an b yields a = −2. resulting in k = 1/6. x > 3 6 18. Therefore:    x + 4. ∞) 16. 1) and (1. we must have f (1) = 5 = lim+ (ax + b) and f (3) = x→1 3a + b = lim+ (3x − 8). we get two equations 5 = a + b and 1 = 3a + b.elsolucionario. x ≤ 3 f (x) = 6 x 2 − . 1. − 5) and ( 5. x≤1 f (x) = −2x + 7. Thus. we must solve for k in the equation 3k + 1 = 2 − 3k. 20. For f (x) to be continuous at the number 3. . nπ + π) for n = 0. 1 < x ≤ 3   3x − 8. [−2. b = 7. we have y − 32 = 4(x − 2) or y = 4x + 24. x>3 19. 2] √ √ 15. −2). (nπ. 32). we must have f (3) = 3k + 1 = lim (2 − kx). 116 CHAPTER 2. Thus. LIMIT OF A FUNCTION With point of tangency (1/2. 2). resulting in the equation y − 2 = (x − 1)/2 or y = (x + 3)/2.01. f (a) = f (4) = 12. f (a + h) = f (1 + h) = −4(h + 1)2 + 6(h + 1) f (a + h) − f (a) = [−4(h + 1)2 + 6(h + 1)] − 2 = −4h2 − 8h − 4 + 6h + 6 − 2 = −4h2 − 2h = h(−4h − 2) f (a + h) − f (a) h(−4h − 2) mtan = lim = lim = −2 h→0 h h→0 h With point of tangency (1. 2). we have y − 12 = 2(x − 4) or y = 2x + 4. we have y − 2 = −2(x − 1) or y = −2x + 4. 24. f (a + h) = f (4 + h) = (h + 4) + 4 h + 4 √ √ √ (h − 8) − 4 h + 4 f (a + h) − f (a) = [(h + 4) + 4 h + 4] − 12 = [(h − 8) + 4 h + 4] · √ (h − 8) − 4 h + 4 (h − 8) − 16(h + 4) 2 h − 16h + 64 − 16h − 64 2 = √ = √ (h − 8) − 4 h + 4 (h − 8) − 4 h + 4 h(h − 32) = √ (h − 8) − 4 h + 4 f (a + h) − f (a) h(h − 32) −32 mtan = lim = lim √ = =2 h→0 h h→0 [(h − 8) − 4 h + 4]h −16 With point of tangency (4. the line that is perpendicular to this line would have a slope of 1/2 and also passes through (1. Thus. √ 22. we choose δ = /2 and so δ = 0. f (a) = f (1) = 2. −2). x→1 . 12). |2x + 5 − 7| = |2x − 2| = 2|x − 1| <  whenever |x − 1| < /2. 23. Finding δ proves that lim (2x + 5) = 7. we have y + 2 = 8(x − 1/2) or y = 8x − 6.005 when  = 0. 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