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Chapter 2
Limit of a Function 2. 2.1 1
Limi Limits ts — An An Info Inform rmal al Appr Approa oac ch 2. lim (x2
1. lim (3x (3x + 2) = 8 x→2
8
− 1) = 3
x→2
• (2, 8) 6
(–2, 3) •
4
-4 -4
3
-2 -2
2
4
4
3. No limit as x
→ 0.
4. lim
x→5
√ x − 1 = 2
4
5 (5, 2) •
-4
-2
2
-5
4
-5
-4
x2 x→1 x
5. lim
5
− 1 = lim (x + 1) = 2 −1 x
6. lim
x→0
→1
5
x2
− 3x = lim (x − 3) = −3 x
x→0
5 (1, 2)
-5
5
-5
5 (0, 3)
-5
-5
73
74
CHAPTE CHAPTER R 2. LIMIT LIMIT OF A FUN FUNCTI CTION ON
7. No limit as x
→ 3.
8. No limit as x
3
→ 0.
3
3
6
-3
3
-3
-3
x3 =0 x→0 x
x4 x→1 x2
9. lim
10. lim
−1 =2 −1
3
3 (1, 2)
(0, 0)
-3
3
-3
3
-3
-3
11. lim f ( f (x) = 3
12. No limit as x
x→0
(0, 3) • 3
→ 2.
3
-3
3
-3
3
-3
-3
13. lim f ( f (x) = 0
14. No limit as x
x→2
3
→ 0.
3
(2, 0) -3
3
-3
3
-3
-3
15. 15.
(a) 1
(b) (b) -1
(c) 2
16.
(a) 0
(b) 3
17.
(a) 2
(b) -1
18. 18.
(a) (a) does doesn’ n’tt exis existt
(c) 3
x→0+
(d) 3
(c) -1 (b) (b) 3
19. Correct Correct 20. Incorre Incorrect; ct; lim
(d) doe doesn’ sn’t exi exist
√ x = 0 4
(d) -1 (c) (c) -2
(d) (d) does doesn’ n’tt exis existt
74
CHAPTE CHAPTER R 2. LIMIT LIMIT OF A FUN FUNCTI CTION ON
7. No limit as x
→ 3.
8. No limit as x
3
→ 0.
3
3
6
-3
3
-3
-3
x3 =0 x→0 x
x4 x→1 x2
9. lim
10. lim
−1 =2 −1
3
3 (1, 2)
(0, 0)
-3
3
-3
3
-3
-3
11. lim f ( f (x) = 3
12. No limit as x
x→0
(0, 3) • 3
→ 2.
3
-3
3
-3
3
-3
-3
13. lim f ( f (x) = 0
14. No limit as x
x→2
3
→ 0.
3
(2, 0) -3
3
-3
3
-3
-3
15. 15.
(a) 1
(b) (b) -1
(c) 2
16.
(a) 0
(b) 3
17.
(a) 2
(b) -1
18. 18.
(a) (a) does doesn’ n’tt exis existt
(c) 3
x→0+
(d) 3
(c) -1 (b) (b) 3
19. Correct Correct 20. Incorre Incorrect; ct; lim
(d) doe doesn’ sn’t exi exist
√ x = 0 4
(d) -1 (c) (c) -2
(d) (d) does doesn’ n’tt exis existt
75
2.1. LIMITS LIMITS — AN INFORMA INFORMAL L APPROA APPROACH CH
√ 1 − x = 0
21. Incorr Incorrect ect;; lim x→1
−
22. Correct Correct 23. Incorre Incorrect; ct; lim+ x = 0
x→0
24. Correct Correct 25. Correct Correct 26. Incorr Incorrect ect;; lim cos−1 x = 0 x→0
−
√ 9 − x
2
27. Incorr Incorrect ect;; lim x→3
−
=0
28. Correct Correct 29. (a) Does not exist 30. (a)
≈ 2. 2 .5
31.
(b) 1
(b) 0 (c )
(c) 3
−1
(d)
−2
(e) 0
(d) Does not exist
32.
4
(e) 0
4 2
-4
4
-4
-2
2
4
2
4
-2 -4
-4
33.
34.
4
4
2
-4
-2
2
2
35.
4
-4
-2
-2
-2
-4
-4
36.
1
0.5
0.5
-1
-0 .5
0.5
1
-0.5
0 .5
-0.5 -0.5 -1
The limit does not exist.
The limit is 0.
(f ) 1 (f ) 0
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76
CHAPTE CHAPTER R 2. LIMIT LIMIT OF A FUN FUNCTI CTION ON
37.
38.
0.5
-0.5
0 .5 -1
-0.5
0.5 -2
-0.5
-3
The limit is 39.
−0.25.
-0.5
The limit is
−3.
40.
0.5
5
-1 -0.5
-2
-5
-3
The limit is 41.
→ 1
x
0 .5
−
f ( f (x) x 1 + f ( f (x)
−2.
−
→
0. 9 3.25536642 1.1 2.79817601
− lim f ( f (x) = −3 x 42. x → 1 0.9
The limit does not exist.
− −
0.99 3.02276607 1.01 2.97775903
0.999 3.00225263 1.001 2.99775260
0.9999 3.00022503 1.0001 2.99977503
−
−
−
−
→1
−
f ( f (x) x 1 + f ( f (x)
1.0536 053605 0516 16 1.1 0.9531 953101 0180 80
→
0.99 1.0050 005033 3359 59 1.01 0.9950 995033 3309 09
0.999 1.0005 000500 0033 33 1.001 0.9995 999500 0033 33
0.9999 1.00005000 1.0001 0.99995000
lim f ( f (x) = 1
x→1
43.
x
→ 0
−
−0.1 −0.01 −0.001 −0.0001 −0.04995835 −0.00499996 −0.00050000 −0.00005000
f ( f (x) x 0 + f ( f (x)
0. 1 0.04995835
→
0.01 0.00499996
0.001 0.00050000
0.0001 0.00005000
lim f ( f (x) = 0
x→0
44. Since Since
1
x
→ 0
f ( f (x)
− cos x is an even even functi function, on, it suffices suffices to consider consider only x → 0 x2
+
0.1 0.4995 499583 8347 47
0.01 0.4999 499995 9583 83
0.001 0.4999 499999 9996 96
0.0001 0.50000000
lim f ( f (x) = 0.5
x→0
x is an even function, it suffices to consider only x 0 + . sin3x sin3x x 0 + 0.1 0.01 0.001 0.0001 f ( f (x) 0.3383 338386 8634 34 0.3333 333383 8334 34 0.3333 333333 3383 83 0.33333334
45. Since Since
→
→
+
.
77
2.2. LIMIT THEOREMS
lim f (x) = 0.33333333
x→0
tan x is an even function, it suffices to consider only x 0 + . x x 0 + 0.1 0.01 0.001 0.0001 f (x) 1.00334672 1.00003333 1.00000033 1.00000000
46. Since
→
→
lim f (x) = 1
x→0
47.
x
→ 4
−
3.9 0.25158234 4.1 0.24845673
f (x) x 4 + f (x)
→
3.99 0.25015645 4.01 0.24984395
3.999 0.25001563 4.001 0.24998438
3.9999 0.25000156 4.0001 0.24999844
lim f (x) = 0.25
x→4
48.
x
→ 3
−
2.9 0.52186477 3.1 0.48008703
f (x) x 3 + f (x)
2.99 0.50209311 3.01 0.49792633
−
→
− lim f (x) = −0.5 x 49. x → 1 0.9
2.999 0.50020843 3.001 0.49979176
2.9999 0.50002083 3.0001 0.49997917
−
−
−
−
−
−
→3
−
f (x) x 1 + f (x)
4.43900000 1.1 5.64100000
→
0.99 4.94039900 1.01 5.06040010
0.999 4.99400400 1.001 5.00600400
0.9999 4.99940004 1.0001 5.00060004
lim f (x) = 5
x→1
50.
−
x
→ −2
f (x) x 2+ f (x)
→ −
−2.1
12.61000000 1.9 11.41000000
−
12.06010000 1.99 11.94010000
lim f (x) = 12
x→−2
2.2
Limit Theorems
1. 15 2. cos π = 3.
−12 4. −3 5. 4
−1
−2.01 −
−2.001
12.00600100 1.999 11.99400100
−
−2.0001
12.00060001 1.9999 11.99940001
−
78
CHAPTER 2. LIMIT OF A FUNCTION
6.
−125
7. 4 8.
−136 9. −8/5 10. does not exist 11. 14 12. 4 13. 28/9 14. lim
x→6
15.
x2 x2
− 6x
− 7x + 6
x(x 6) x 6 = lim = x→6 (x 1)(x 6) x→6 x 1 5
= lim
−
−
−
−
−1
16. 16 17.
√ 7
18. 3 19. does not exist 20. 16 y 2 25 = lim (y y→−5 y + 5 y→−5
−
21. lim
− 5) = −10
u2
− 5u − 24 = lim (u + 3) = 11 u u u−8 x −1 (x − 1)(x + x + 1) 23. lim = lim = lim (x + x + 1) = 3 x x x x−1 x−1 t +1 (t + 1)(t − t + 1) t − t + 1 3 24. lim = lim = lim = − t t t −1 t (t + 1)(t − 1) t−1 2 (x − 2)(x + 5) 8(15) 25. lim = = 60 x x−8 2 2x + 6 2(x + 3) 1 1 26. lim = lim = lim = − x 4x − 36 x 4(x + 3)(x − 3) x 2(x − 3) 12 22. lim
→8
→8
3
2
2
→1
→1
→1
3
→−1
2
2
2
→−1
→−1
→10
→−3
2
→−3
→−3
x3 + 3x2 10x x(x + 5)(x = lim x→2 x→2 x 2 x 2
27. lim
−
−
−
− 2) = lim x(x + 5) = 14 x→2
2x2 + 3x 9 (2x 3)(x + 3) 2(x 1.5)(x + 3) = lim = lim = lim 2(x + 3) = 9 x→1.5 x→1.5 x→1.5 x→1.5 x 1.5 x 1.5 x 1.5
28. lim
−
−
−
−
−
−
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79
2.2. LIMIT THEOREMS
t3 2t + 1 (t 1)(t2 + t 1) t2 + t 1 1 29. lim 3 = lim = lim 2 = 2 2 t→1 t + t 2 t→1 (t 1)(t + 2t + 2) t→1 t + 2t + 2 5
−
− −
−
−
−
x3 1 1 = lim = 4 3 x→0 x + 2x x→0 x + 2 2
30. lim x3 (x4 + 2x3 )−1 = lim x→0
31. lim
x→0+
(x + 2)(x5 1)3 2( 1) = = 16 ( x + 4)2
−
√
−18
−
√ − 6 = −2√ 2 √ −8 = 4√ 2
√
32. lim x x + 4 3 x x→−2
x2 + 3x 33. lim x→0 x 34. lim
x→2
−1 + 1
1
− 2 −
x
3
x
6 2 x + 2x
x2 + 3x = lim (x + 3) = 3 x→0 x→0 x
= lim
−8
= lim
− − 1
2 (x 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 x→2
x
6 2)(x + 4)
−
−
−
−
− −
35. does not exist 36.
−2
10
or
−1024
37. 2
√ √
2 2 38. 3 4 39. lim
h→4
h h2 h + 5 h
− 16 −4
2
= lim
h→4
h 128 (h2 + 8h + 16) = h + 5 3
40. 16 41. lim x→0
−
− 5
x3 64x = lim x2 + 2x x→0
−
−100, 000 43. a − 2ab + b √ 44. lim u x + 2xu + 1 =
−
x2 64 = x + 2
5
−2
42.
2
2
2 2
x→−1
(8 + h)2 45. lim h→0 h
x→−1
− 64 = lim 16h + h
1 [(1 + h)3 h→0 h
46. lim
√ u − 2u + 1 = lim x 2
lim
h
h→0
2
− 1] = hlim (h →0
→−1
2
−
= lim (16 + h) = 16 h→0
+ 3h + 3) = 3
(u
1)2 = u
| − 1|
80
CHAPTER 2. LIMIT OF A FUNCTION
1 47. lim h→0 h
1 x + h
−
1 1 = lim h→0 h x = lim
x
− (x + h)
(x + h)x 1 1 = 2 x + hx x2
= lim
h→0
−h
hx(x + h)
− − √ x + h − √ x √ x + h − √ x √ x + h + √ x √ x + h + √ x lim = lim h h h h (x + h) − x 1 = lim √ √ = lim √ 1 √ = √ h→0
48.
→0
→0
h→0
h( x + h +
x)
h→0
x + h +
2 x
x
√ t − 1 √ t − 1 √ t + 1 1 1 √ 49. lim = lim = lim √ = t t t t−1 t − 1 t + 1 2 t + 1 √ u + 4 − 3 √ u + 4 − 3 √ u + 4 + 3 √ u + 4 + 3 50. lim = lim u u u−5 u−5 u−5 1 1 √ = lim = lim √ = u u 6 (u − 5)( u + 4 + 3) u + 4 + 3 √ 25 + v − 5 √ 25 + v − 5 √ 25 + v + 5 √ 1 + v + 1 √ 1 + v + 1 √ 25 + v + 5 51. lim √ = lim √ v v 1 + v − 1 1 + v − 1 √ v 1 + v + 1 1 = lim √ = v v 25 + v + 5 5 √ √ √ 4 − x + 15 4 − x + 15 4 + x + 15 √ 52. lim = lim x x x −1 x −1 4 + x + 15 1−x √ = lim x (x + 1)(x − 1)(4 + x + 15) −(x − 1) √ = lim x (x + 1)(x − 1)(4 + x + 15) −1√ 1 = lim = − x 16 (x + 1)(4 + x + 15) →1
→1
→5
→1
→5
→5
→0
→5
→0
→0
→1
2
→1
→1
→1
→1
53. 32 54. 64 1 55. 2 56.
4 = 2
√ 2
57. does not exist 58. 8 59. 8a 3 60. 2
2
81
2.3. CONTINUITY
x100 1 x100 1 61. (a) lim 2 = lim x→1 x x→1 (x + 1)(x 1 1) 100 1 x 1 1 = lim = 100 = 50 x→1 x + 1 x 1 2 x50 1 x50 1 x 50 + 1 (b) lim = lim x→1 x x→1 x 1 1 x50 + 1 x100 1 1 1 = lim = 100 = 50 50 x→1 x 1 x +1 2 (x100 1)2 x100 1 x 100 1 (c) lim = lim = 100 100 = 10, 000 x→1 (x x→1 x 1)2 1 x 1
− −
−
−
− −
·
− −
·
− · − − · − − · −
− −
2x 62. (a) lim = 2 lim x→0 sin x x→0
·
− −
·
lim 1
1 sin x x
=2
·
x→0
sin x x→0 x
=2
lim
sin2 x sin x sin x = lim =1 1 =1 2 2 x→0 x→0 x x→0 x x x 8x2 sin x sin x sin x (c) lim = lim 8x = lim 8x lim = x→0 x→0 x→0 x→0 x x x
(b) lim
1
− cos
2
x
= lim
−
63. lim sin x = lim x→0
64.
x→0
lim [2f (x)
x→2
2 lim f (x) x→2
−
· x
− 5] = xlim (x + 3) →2
− lim 5 = 20 x
−
·
·
2f (x) 5 = 5 4 = 20 x + 3
−
→2
x→2
20 + 5 = 12.5 2
Continuity
1. Continuous everywhere 2. Continuous everywhere 3. Discontinuous at 3 and 6 4. Discontinuous at
−1 and 1
nπ , for n = 0, 1, 2, . . . 2 π 6. Discontinuous at 3 and + nπ, for n an integer 2 5. Discontinuous at
−
7. Discontinuous at 2 8. Discontinuous at 0
·
sin x sin x = lim x lim =0 1= 0 → 0 → 0 x x x x
lim f (x) =
2.3
·
·
−1
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82
CHAPTER 2. LIMIT OF A FUNCTION
9. Continuous everywhere 10. Discontinuous at x < 0 and
1 2
11. Discontinuous at e −2 12. Discontinuous at 0 13. (a) yes (b) yes 14. (a) no (b) yes 15. (a) yes (b) yes 16. (a) yes (b) yes 17. (a) no (b) no 18. (a) yes (b) yes 19. (a) yes (b) no 20. (a) no (b) no 21. Solving 2 + sec x = 0, we obtain cos x = 12 , so x = is discontinuous on ( , ) and on [ π2 , 32π ].
−
−∞ ∞
2π 3
+ 2nπ or x =
22. Since sin x1 is discontinuous only at x = 0, it is continuous on [ π1 , [ −π2 , π2 ].
4π 3
+ 2nπ. Thus, f (x)
∞) and discontinuous on
23. Since f (x) is discontinuous only at x = 2, it is discontinuous on [ 1, 3] and continuous on (2, 4].
−
24. Since f (x) is defined and continuous exactly on (1, 5], it is continuous on [2 , 4] and discontinuous on [1, 5]. 25. Since lim f (x) = 4m and lim f (x) = 16, we have 4m = 16 and m = 4. x→4
x→4+
−
26. Since lim f (x) = lim x→2
(x
x→2
− 2)(x + 2) = 4 we have f (2) = m and m = 4. x−2
27. Since lim f (x) = 3m, lim+ f (x) = 3, and f (3) = n, we have 3m = 3 = n, so m = 1 and x→3
−
n = 3.
28. Since lim f (x) = m x→1
−
x→3
− n, xlim
→1+
f (x) = 2m + n, and f (1) = 5, we have m
2m + n = 5. Adding, we obtain 3m = 10, so m = 10/3 and n =
−5/3.
− n = 5 and
83
2.3. CONTINUITY
n , n an integer 2
29. Discontinuous at 3
-3
3
-3
30. Discontinuous at every integer 3
-3
3
-3
√ √ − 3) √ − √ − √ − = lim ( x + 3) = 6, define f (9) = 6. x x −1 (x + 1)(x − 1) 32. Since lim = lim = lim (x + 1) = 2, define f (1) = 2. x x x −1 x x −1 √ 3 π π 2π x 9 ( x + 3)( x 31. Since lim = lim x→9 x 3 x→9 x 3 4
2
→9
2
2
→1
33.
lim sin(2x +
x→π/6
34. lim2 x→π
35.
36.
2
2
→1
3
) = sin
√ cos x = cos
lim
x→π2
lim sin(cos x) = sin
x→π/2
lim (2x +
x→π/ 6
√ x
37. lim cos t→π
38. lim tan t→0
39. lim
t→π
−π −π
x→π/2
2
= cos lim
(t
t→π
3
=
2
−1
π ) = sin 0 = 0 2
lim cos x = 1 + cos(cos
x→π/ 2
− π)(t + π) t−π
π ) = 1 + cos 0 = 2 2
= cos 2π = 1
πt πt π π = tan lim = tan lim = tan = 2 t→0 t(t + 3) t→0 t + 3 t + 3t 3
√ t − π + cos
2
t =
√
cos2 π =
√ 1 = 1
40. lim (4t + sin 2πt)3 = lim (4t + sin 2πt) t→1
) = sin
lim cos x = sin(cos
x→π/2
t2 t
3
= cos π =
lim [1 + cos(cos x)] = 1 + cos
→1
t→1
3
= (4 + sin 2π)3 = 4 3 = 64
√ 3
84 41.
CHAPTER 2. LIMIT OF A FUNCTION
−1
lim sin
x→−3
x + 3 x + 3 −1 = sin lim x→−3 x2 + 4x + 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
42. lim ecos 3x = e
lim cos 3x
◦
−
−
= e cos 3π = e −1
x→π
x→π
43. Since (f g)(x) =
1 √ x + , f ◦ g is continuous for x + 3 > 0 or on (−3, ∞). 3
5(x 2)2 5(x 2)2 5(x 2)2 = = , we see that f g is continuous (x 2)2 1 x2 4x + 3 (x 1)(x 3) for x = 1 and x = 3 or on ( , 1) (1, 3) (3, ).
44. Since (f g)(x) =
◦
−
−
−
− − − ∞ ∪ ∪ ∞
−
◦
−
45. f (1) = −1, f (5) = 15. By the Intermediate Value Theorem, since −1 ≤ 8 ≤ 15, there exists c ∈ [1, 5] such that c − 2c = 8. Setting c − 2c − 8 = 0 gives us (c − 4)(c + 2) = 0 or c = −2, 4. 2
2
On [1, 5], c = 4.
46. f ( 2) = 3, f (3) = 13. By the Intermediate Value Theorem, since 3
− ≤ 6 ≤ 13, there exists −1 ± 1 − 4(1)(−5) = c ∈ [ −2, 3] such that c + c +1 = 6. Setting c + c − 5 = 0 gives us c = 2 −1 ± √ 21 . On [−2, 3], c = −1 + √ 21 . 2 2 47. f (−2) = −3, f (2) = 5. By the Intermediate Value Theorem, since −3 ≤ 1 ≤ 5, there exists c ∈ [ −2,√ 2] such that c − 2x + 1 = 1. Setting c − 2c = 0 gives us c(c − 2) = 0. On [ −2, 2], c = 0, ± 2. 48. f (0) = 10, f (1) = 5. By the Intermediate Value Theorem, since 5 ≤ 8 ≤ 10, there exists 10 5 1 1 1 c ∈ [0, 1] such that = 8. Setting c + 1 = or c − = 0 gives us (c + )(c − ) = 0 c +1 4 4 2 2 1 1 or c = ± . On [0, 1], c = . 2 2 49. Since f (0) = −7, f (3) = 242, and −7 ≤ 50 ≤ 242, then by the Intermediate Value Theorem there exists c ∈ [0, 3] such that f (c) = 50. 50. Since f (a) > g(a), then (f − g)(a) > 0. Since f (b) < g(b), then (f − g)(b) < 0. By the corollary to the Intermediate Value Theorem, there exists c ∈ (a, b) such that (f − g)(c) = 0. 2
2
3
3
2
2
2
2
Then f (c) = g(c).
51. The equation will have a solution on (0, 1) if f (x) = 2x7 + x 1 is 0 on (0, 1). Since f (0) = and f (1) = 2, then by the Intermediate Value Theorem f (c) = 0 for some c (0, 1).
−
∈
−1
x2 + 1 x 4 + 1 1 1 1 2 + . Then f (0) = > 0 and f (1) = < 0. Thus, by the x + 3 x 4 3 4 2 3 corollary to the Intermediate Value Theorem, f (c) = 0 for some c between 0 and 1, and hence between 3 and 4.
52. Let f (x) =
−
−
−
−
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85
2.3. CONTINUITY
53. Let f (x) = e −x ln x. Then f (1) = e −1 ln 1 = e −1 > 0 and f (2) = e −2 ln2 < 0. Thus, by the corollary to the Intermediate Value Theorem, f (c) = 0 for some c (1, 2).
−
−
∈
−
π sin 1 2 2 = 2 , sin π = 0, and 0 54. Since , then by the Intermediate Value Theorem, π π π 2 π 2 π sin x 1 there exists c between and π such that = . 2 x 2
≤ ≤
3
55. In [ 2, 1] the zero is approximately 1.21. In [ 1, 0] the zero is approximately 0.64. In [1, 2] the zero is approximately 1 .34.
− − −
−
−
-3
3
-3
3
56. In [0, 1] the zero is approximately 0 .75. -3
3
-3
57. We want to solve f (x) = x5 + 2x 7 = 50 or x5 + 2x 57 = 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. Applying the bisection method on [2, 3], we find c 2.21.
−
58. Applying the bisection method to f (x) = 2x7
−
≈ + x − 1 on [0, 1], we find c ≈ 0.75.
59. In the solution of Problem 52 we saw that there is a zero in [0, 1]. Applying the bisection method on this interval, we find c 0.78.
≈
60. (a) If h is the height of the cylinder, then the volume is given by V = πr 2 h and the surface S area is S = 2πr 2 + 2πrh. Solving the latter equation for h, we obtain h = r. 2πr 1 Substituting into the formula for V , we find V = Sr πr 3 or 2πr 3 Sr + 2V = 0. 2 (b)
−
−
−
5000
10
20
-5000
(c) From the graph, we observe zeros in [3, 4] and [14, 15]. The bisection method gives 1800 r 3.48 ft and r 14.91 ft. The corresponding values of h are h = r 78.84 ft 2πr 1800 and h = r 4.29 ft. 2πr
≈
≈ − ≈
− ≈
86
CHAPTER 2. LIMIT OF A FUNCTION
61. Since f and g are continuous at a, then lim f (x) = f (a) and lim g(x) = g(a). 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) Thus, f + g is continuous at a. 62. Since f and g are continuous at a, then lim f (x) = f (a) and lim g(x) = g(a). 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) = 0)
Thus, f /g is continuous at a. 63. f g will be discontinuous whenever cos x is an integer. In the interval [0, 2π), this will be the case whenever x = 0, π /2, π , or 3π/2. Thus, f g will be discontinuous for x = nπ/2, n an integer.
◦
64. (f g)(x) =
◦
◦
|
x + 1|, x < 0 |x − 1|, x ≥ 0
is continuous at x = 0.
3
-3
3
-3
(g f )(x) = x
◦
| | − 1 is continuous at x = 0.
3
-3
3
-3
65. (a) For any real a, lim f (x) does not exist since f takes on the values 0 and 1 arbitrarily x→a close to any real number. Therefore, the Dirichlet function is discontinuous at every real number. (b) The graph consists of infinitely many points on each of the lines y = 0 and y = 1. In fact, between any two real numbers, 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. (c) Let r be a positive rational number. If x is rational, then x + r is rational so that f (x+r) = 1 = f (x). If x is irrational, then x +r is irrational so that f (x+r) = 0 = f (x).
2.4
Trigonometric Limits sin3t 1 sin3t 3 = lim = t→0 2t 2 t→0 t 2
1. lim
87
2.4. TRIGONOMETRIC LIMITS
sin( 4t) = t→0 t
2. lim
−
−4
sin x 0 = =0 x→0 4 + cos x 4+1
3. lim
1 + sin x 1+0 1 = = x→0 1 + cos x 1+1 2
4. lim
cos2x =1 x→0 cos3x
5. lim
tan x 1 6. lim = lim x→0 3x 3 x→0
sin x 1 1 1 = (1 1) = x cos x 3 3
·
1 7. lim = lim t→0 t sec t csc4t t→0
·
sin4t cos t = 4 1 = 4 t
·
·
t cos2t 1 8. lim (5t cot2t) = 5 lim = 5 lim cos2t t→0 t→0 sin2t t→0 (sin2t)/t 1 1 5 = 5 lim cos2t lim = 5 1 = t→0 t→0 (sin2t)/t 2 2
2sin2 t 9. lim = 2 lim t→0 t cos2 t t→0
·
· ·
sin t sin t = 2 1 0 = 0 t cos2 t
·
· ·
sin2 (t/2) sin(t/2) t sin(t/2) 10. lim = lim t→0 t→0 sin t t sin t sin(t/2) sin(t/2) 1 0 = lim lim = =0 t→0 t→0 (sin t)/t t 2 1
·
·
sin2 6t 11. lim = lim t→0 t→0 t2
sin6t t
·
2
= 6 2 = 36
t3 t2 12. lim = lim t = lim t t→0 sin2 3t t→0 t→0 sin2 3t
1 1 lim = 0 =0 t→0 [(sin 3t)/t]2 32
·
sin(x 1) 1 sin(x 1) 1 = lim = x→1 2x 2 2 x→1 x 1 2
13. lim
14. lim
x→2π
− − x − 2π = sin x
− −
lim
x→2π
x 2π = lim sin(x 2π) x→2π sin(x
cos x does not exist. x→0 x
− −
15. lim 16.
1 + sin θ does not exist. cos θ θ →π/ 2 lim
cos(3x π/2) sin3x = lim =3 x→0 x→0 x x
17. lim
−
−
1 2π)/(x
− 2π) = 1
www.elsolucionario.org
88
CHAPTER 2. LIMIT OF A FUNCTION
sin(5x + 10) 5 1 = x→−2 4x + 8 5 4
18. lim
sin(5x + 10) 5 sin(5x + 10) 5 = lim = →−2 x + 2 4 x→−2 5x + 10 4
· · xlim
sin3t 19. lim = lim t→0 sin7t t→0
· · √ ·
sin3t t = t sin7t
·
t→0
→0+
t
1 1 3 = 3 = t→0 (sin7t)/t 7 7 lim
·
sin2t sin2t t = lim t→0 sin3t t→0 t sin3t sin2t 1 1 2 = lim lim = 2 = t→0 t→0 (sin3t)/t t 3 3
20. lim sin2t csc3t = lim
21. lim
sin3t t→0 t lim
sin t = lim t→0+ t
√
√ · t
sin t = t
lim
t
→0+
t
lim
t
→0+
sin t = 0 1 = 0 t
√ √ 1 − cos t √ = lim 1 − cos u = 0. 22. Letting u = t, we have lim t
t→0
23. lim
t→0
t2
− 5t sin t = lim 1 − 5 t t 2
→0
u
u→0
sin t t
cos4t 1 = =1 t→0 cos8t 1
= 1
− 5 = −4
24. lim
√
√
(x + 2 sin x)2 x2 + 4x sin x + 4 sin x 25. lim = lim x x x→0+ x→0+ 4 sin x = lim x + 4 sin x + = 0 + 0 + 4 = 4 + x x→0
√
26. lim
x→0
(1
− cos x) x
2
=
cos x 1 27. lim = x→0 cos2 x 1
− −
lim (1
x→0
− cos x)
cos x lim x→0 cos x
−1 −1
sin x + tan x sin x 28. lim = lim + x→0 x→0 x x
lim
x→0
1
− cos x x
= 0 0 = 0
·
1 1 1 lim = 1 = x→0 cos x + 1 2 2
sin x 1 x cos x
·
·
= 1 + (1 1) = 2
·
sin5x2 sin5u = lim = 5. 2 x→0 u→0 x u
29. Letting u = x 2 , we have lim t2 30. lim = lim t→0 1 cos t t→0
−
t2 1 + cos t t2 = lim (1 + cos t) t→0 sin2 t 1 cos t 1 + cos t
·
−
1 = lim t→0 (sin t)/t
2
·
lim (1 + cos t) = 12 2 = 2
t→0
·
sin(x 2) sin(x 2) as lim . 2 x→2 x + 2x x→2 (x 8 2)(x + 4)
31. First, rewrite lim Letting u = x
− −
− 2, we get ulim
→0
−
−
sin u 1 1 1 = 1 = . u u + 6 6 6
·
·
89
2.4. TRIGONOMETRIC LIMITS
x2 9 (x 3)(x + 3) 32. First, rewrite lim as lim . Letting u = x x→3 sin(x x→3 3) sin(x 3)
− −
−
−
− 3:
u 1 1 lim (u + 6) = lim (u + 6) = 6= 6 u→0 sin u u→0 (sin u)/u 1 2sin4x + 1 33. lim x→0 x 34. lim
x→0
4x2
·
− cos x = lim
x→0
− 2sin x = lim x
x→0
4x
−
·
2sin4x 1 + x
− cos x
− 2 = −2
2sin x = 0 x
x
·
= 8 + 0 = 8
1 + tan x , producing: 1 + tan x 1 tan x 1 tan x 1 + tan x lim = lim sin x x→π/4 cos x sin x 1 + tan x x→π/ 4 cos x 1 tan2 x = lim sin x)(1 + tan x) x→π/4 (cos x
35. Start by multiplying the function by
−
−
−
−
−
·
−
Focusing first on the denominator, we multiply out and simplify: (cos x
− sin x)(1 + tan x) = cos x + cos x tan x − sin x − sin x tan x sin x cos x sin x − − = cos x + cos x sin x sin x cos x cos x cos x sin x cos x − sin x = cos x − = 2
2
2
cos x cos x Substituting this result back into the function, we get: 1
2
− − − − − − −
cos x (cos x x sin2 x sin2 x cos x cos x cos x cos2 x = = cos2 x sin2 x cos2 x cos2 x sin2 x 1 = cos x cos2 x sin2 x Finally, returning to the limit, we have: 1 tan x 1 1 lim = lim = = 2 sin x x→π/4 cos x x→π/4 cos x 2/2
−
− tan
x = (1 sin x)(1 + tan x)
2
tan x)
cos2
−
−
2
36. Using the trigonometric identity cos 2x = cos2 x cos2x = lim sin x x→π/4 x→π/4 cos x lim
−
√
√
−
− sin x, we have: cos x − sin x cos x − sin x 2
2
(cos x + sin x)(cos x cos x sin x x→π/ 4
= lim
− √
= lim (cos x + sin x) = x→π/ 4
sin2 x cos x sin2 x 1 = cos x
2
− sin x)
90
CHAPTER 2. LIMIT OF A FUNCTION
π f + h 4 37. lim h→0
π f 4
−
sin
π 4
f
− − √
√
√
h→0
sin
= lim h→0 h 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
√
38. lim
π + h 4
− √
−
π + h 6
π 6
− f
cos
√
− − π + h 6
cos
π 6
= lim h→0 h 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
−
√
= lim
h→0
39. Since
−
√ 3 2
− − √
cos h h
− 1 − 1 2
sin h h
=
√ 3 2
· 0 − 12 · 1 = −12
−1 ≤ sin 1x ≤ 1, then −|x| ≤ x sin 1x ≤ |x|. Since xlim (−|x|) = 0 and xlim |x| = 0, then
1 by the Squeeze Theorem, lim x sin = 0. x→0 x π π 40. Since 1 cos 1, then x2 x 2 cos x x π 2 by the Squeeze Theorem, lim x cos = 0. x→0 x
− ≤
≤
− ≤
→0
2
2
≤ x . Since xlim −x →0
41. For both limits, we use the result from Problem 39, lim x sin x→0
→0
= 0 and lim x2 = 0, then x→0
1 = 0: x
1 1 1 (a) lim x3 sin = lim x2 x sin = lim x2 lim x sin = 0 0 = 0 → 0 → 0 → 0 x→0 x x x x x x 1 1 1 (b) lim x2 sin2 = lim x sin lim x sin = 0 0 = 0 x→0 x→0 x→0 x x x
·
·
·
·
42. f (x) B means that B 0 and therefore B f (x) B. Thus, Bx 2 x 2 f (x) Bx2 in that interval. Since lim ( Bx 2 ) = 0 and lim Bx 2 = 0, then by the Squeeze Theorem,
|
|≤
≥
x→0
lim x2 f (x) = 0.
− ≤
−
≤
−
≤
≤
x→0
x→0
2
43. Since lim (2x
− 1) = 3 and xlim (x − 2x +3) = 3, then by the Squeeze Theorem, lim f (x) = 3. x 44. Since |f (x) − 1| ≤ x , then f (x) − 1 ≤ x , or f (x) ≤ x + 1 when f (x) − 1 > 0. However, f (x) ≤ x +1 is in fact true for all x, since x ≥ 0 for all x. Similarly, we have −x ≤ f (x) − 1, or −x + 1 ≤ f (x) for all x. Since lim (−x +1) = 1 and lim (x +1) = 1, then by the Squeeze x x x→2
→2
→2
2
2
2
2
2
2
2
2
→0
Theorem, lim f (x) = 1. x→0
2
→0
www.elsolucionario.org
91
2.4. TRIGONOMETRIC LIMITS
− π4 . Thus, x = t + π4 and we have the following substitutions: √ 2 √ 2 π π π
45. Let t = x
sin x = sin(t +
sin x
−
) = sin t cos + cos t sin = 4 4 4 π π π cos x = cos(t + ) = cos t cos sin t sin = 4 4 4 2 2 2 cos x = sin t + cos t cos t 2 2 2
√
√
−
√ −
sin x cos x With these substitutions, lim = lim t→0 (π/4) x→π/4 x
−
46. Let t = x
−
−
sin t +
2
√ 2
2
cos t
√ 2
cos t sin t 2 2 2 sin t = 2sin t 2
−
√
√
√ 2sin t √ = 2. t
− π. Thus, x = t + π. Substituting, we get: x−π t lim = lim = lim x→π
t tan2x t→0 tan(2t + 2π) t→0 tan2t 1 1 1 1 = lim = lim = = 1 sin2t t→0 t→0 tan2t 1 2 2 cos2t t t
·
·
47. Let t = π
− (π/x). Therefore π /x = π − t and sin(π/x) = sin(π − t) = sin t. In addition, we t can derive x − 1 = , giving us: π−t sin(π/x) (sin t)(π − t) sin t · lim lim = lim = lim (π − t) = 1 · π = π x t t t x−1 t t →1
48. Let t =
π 2
→0
→0
→0
− πx . Substituting in the same way as in Problem 47, we get: cos(π/x) (sin t)(π − 2t) sin t π − 2t π · lim = lim = lim lim = x t t t 4t 4 4 x−2 t →2
→0
→0
→0
sin x = 1 = f (0). x→0 x
49. f is continuous at x = 0 because lim 50. Since x =
| |
sin x x, x > 0 , knowing that lim = 1 means: x, x < 0 x→0 x
−
sin x sin x = lim =1 x x→0+ x→0+ x sin x sin( x) lim = lim = lim x x x→0 x→0 x→0 lim
−
Since lim+ x→0
|| ||
−
−
−
− sin x = − x
lim
x→0
sin x sin x sin x = lim , then lim does not exist. x→0 x x x x→0
| |
−
||
||
−
sin x = x
−1
92
CHAPTER 2. LIMIT OF A FUNCTION
2.5
Limits that Involve Infinity
1. 5.
2. 6.
−∞ ∞
3. 7.
∞ −∞
4. 8.
∞ ∞
−∞ −∞
x2 3x 1 3/x 1 = lim = 2 2 x→∞ 4x + 5 x→∞ 4 + 5/x 4
−
9. lim
−
x2 1 = lim = − 2 2 x→∞ 1 + x x→∞ 1/x + 1/x4
10. lim
∞
11. 5 12.
lim
x→−∞
√
√
6 1 + 5 = 0 3 x x
− √ √
√ − √ −14 √ 1/ x + 7 7
8 x (8/ x) 1 13. lim = lim = x→∞ 1 + 4 x x→∞ (1/ x) + 4
√ √
1+7 3x 14. lim = lim x→−∞ x→−∞ 23x 15. lim
x→∞
16. lim
x→∞
17. lim
x→∞
18.
lim
3x x + 2
x 1 2x + 6
x 3x + 1
3
x→∞
= lim
x→∞
= lim
3 + 2/x = 6 8/x
1 2 = 2 2
2x 1 = lim 7 16x x→−∞
x→∞
3
2 1/x = 7/x 16
x2 + 1 = lim x x→∞
x2 + 5x
x = lim
x→∞
x→∞
21. lim cos 22.
lim sin
x→−∞
5 5 = cos lim x→∞ x x πx 3 6x
−
1 1/x = 3 2 + 6/x
1 3 + 1/x
= lim
x→∞
2
3 1 + 2/x
3
4x2 + 1 2x2 + x
3x + 2 = lim 6x 8 x→∞
19. lim x
20. lim
=
2
− − − − − √ − − − − − − − − − √ − √ · √ √ − − · √ √ √
x→−∞
x→∞
3
= lim sin x→−∞
x2 + 1
4 + 1/x2 2 + 1/x
2 = 16
3
x + x +
1 5 = 2 2
3
=
1 3 8 2 = 3 3
·
1 2
x2 + 1 = lim x2 + 1 x→∞ x +
−1 √ x
2
+1
=
−12
x2 + 5x + x x2 + 5x + x 5x 5 5 = lim = 2 x2 + 5x + x x→∞ 1 + 5/x + 1 x2 + 5x
x
= 1
π = sin 3/x 6
−
lim
x→−∞
π 3/x 6
−
=0
93
2.5. LIMITS THAT INVOLVE INFINITY
23.
lim sin−1
x→−∞
| | − | | | |
√
x 4x2 + 1
= lim sin−1 x→−∞
= sin
24. lim ln x→∞
x x + 8
x x 4 + 1/x2
−1
1 4 + 1/x2
lim
x→−∞
x→∞
4x 1 + 4x + 1 x x 25. Start with = . From this, 2 x +1 1 + 1/x2 4 + 1/x lim f (x) = lim = 4. x→∞ x→∞ 1 + 1/x2
√ 9x
− − | | | | √ − − || || √ √ √ − − −√ | | − − √ − − −
x→−∞
= sin−1
π 6
x→−∞
lim f (x) = lim
x→−∞
4 1/x = 1 + 1/x2
− −
−4 and
√ 9 −5 =
x→−∞
2x 1 + 2x + 1 x x 27. Start with = . From this, lim f (x) = lim x→−∞ x→−∞ 3x2 + 1 3 + 1/x2 2 3 2 + 1/x 2 2 3 and lim f (x) = lim = = . x→∞ x→∞ 3 3 3 3 + 1/x2
√
9 + 6/x2 = 5 + 1/x
−
2 1/x = 3 + 1/x2
− −
6 3 + 2 5x + 6x + 3 x x 28. Start with = . From this, lim f (x) = lim x→−∞ x→−∞ x4 + x2 + 1 1 + 1/x2 + 1/x4 2 5 5 + 6/x + 3/x 5 = 5 and lim f (x) = lim = = 5. x→∞ x→∞ 1 1 1 + 1/x2 + 1/x4 5+
2
x
lim e
ex e−x 29. lim x = x→−∞ e + e−x
lim e
x→−∞
lim
x→−∞
−√ −
−x
x→−∞
ex
+
0 =
−x
lim e
0+
x→−∞
−x
= lim
x→−∞
ex e−x lim x = x→∞ e + e−x
−
−e
e−x
lim ex
x→∞
= lim
−
x→−∞
lim e−x
x→∞ x→∞
ex = lim 1 = 1 x→∞ ex x→∞
= lim
−x
lim e
x→−∞
lim e−x
x→−∞
−1 = −1
lim ex + lim e−x
x→∞
−
=
lim ex
x→∞
−
0
lim ex + 0
x→∞
= ln 1 = 0
x→−∞
+6 = 5x 1
x x 4 + 1/x2
1 = 2
lim f (x) = lim
9 + 6/x2 26. Start with . From this, 5x 1 x x 9 + 6/x2 3 9 3 and lim f (x) = lim = = . x→∞ x→∞ 5 5 1/x 5 5 2
− − −
= lim sin−1
1 1 = ln lim x→∞ 1 + 8/x 1 + 8/x
= lim ln
√
− √ 23 =
5 6/x + 3/x2 = 1 + 1/x2 + 1/x4
− −
www.elsolucionario.org
94
30.
CHAPTER 2. LIMIT OF A FUNCTION
lim
x→−∞
2e−x 1+ x = 1 + e + e−x
lim 2e−x
0+ lim
x→∞
31.
lim
x→−∞
lim
x→∞
32.
lim
x→−∞
x→−∞
=1+
lim 2e−x
x→−∞
lim e−x
x→−∞
2e−x 1+ x = 1 + e + e−x
lim
x→−∞
ex
x→∞
+
−x
lim e
x→−∞
2e−x = 1 + lim 2 = 3 x→−∞ e−x x→−∞ lim 2e−x
x→∞
lim ex + lim e−x
x→∞
x→∞
→−∞
→∞
x
x→−∞ x x 5 + 1/x = lim = 5 x→−∞ 1 1 4x + x 1 5x 1 5 1/x = lim = lim = lim =5 x→∞ x→∞ x→∞ x x 1 x→−∞
|4x| + |x − | x
−
−
−
-5
5
Vertical asymptote: Horizontal asymptote:
none y = 0
Vertical asymptote: Horizontal asymptote:
none y = 0
Vertical asymptote: Horizontal asymptote:
x = 1 none
5
-5
5
-5
5
35.
0 =1 x→∞ ex
= 1 + lim
→−∞
-5
34.
|x − 5| = lim −x + 5 = lim −1 + 5/x = −1 x x x−5 x−5 1 − 5/x |x − 5| = lim x − 5 = 1 x x−5 x−5 |4x| + |x − 1| = lim −4x − (x − 1) = lim −5x + 1
5
33.
= 1 + lim
−
lim
-5
5
-5
−
−
95
2.5. LIMITS THAT INVOLVE INFINITY
5
36.
-5
5
Vertical asymptote: Horizontal asymptote:
x = 1 y = 1
Vertical asymptote: Horizontal asymptote:
x = 0, x = 2 y = 0
Vertical asymptote: Horizontal asymptote:
none y = 4
Vertical asymptote: Horizontal asymptote:
x = 1 y = 1
Vertical asymptote: Horizontal asymptote:
x = 0 y = 1
Vertical asymptote: Horizontal asymptote:
none y = 1, y = 1
Vertical asymptote: Horizontal asymptote:
x = y =
−
-5
5
37.
-5
5
-5
5
38.
-5
5 -5
5
39.
-5
5
-5
10
10
40. -10
−
10
41.
-10
10
−
-10
10
42.
-10
10
-10
−1, x = 1 −1, y = 1
96 43. 44. 45. 46.
CHAPTER 2. LIMIT OF A FUNCTION
(a) 2
(b)
−∞ (c) 0 (d) 2 (a) ∞ (b) ∞ (c) 1 (d) 3 (a) −∞ (b) −3/2 (c) ∞ (d) 0 (a) ∞ (b) −∞ (c) 0 (d) 0 5
47.
-5
5
-5
5
48.
-5
5
-5
5
49.
-5
5
-5
5
50.
-5
5
-5
·
3 3 3/x sin3/x 51. lim x sin = lim x sin = lim x(3/x) x→∞ x→∞ x x→∞ x 3/x 3/x 3 sin3/x sin3/x = lim x lim = lim 3 lim x→∞ x→∞ 3/x x→∞ x→∞ 3/x x At this point, we substitute t = 3/x, resulting in:
lim 3
x→∞
52. lim v→c
−
m0 = lim 1 v2 /c2 v →c
−
−
sin3/x sin t lim = 3 lim =3 x→∞ 3/x t→0 t
√ 1m− 1 = vlimc 0
→
−
m0 ; so as v 0
→ c
−
, m
→ ∞.
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97
2.5. LIMITS THAT INVOLVE INFINITY
53.
x
10 1.99986667
→ ∞ f (x)
lim x2 sin
x→∞
54.
x
f (x)
lim
x→∞
1000 2.00000000
10000 2.00000000
100 0.99501240
1000 0.99950012
10000 0.99995000
2 =2 x2 10 0.95114995
→ ∞
100 1.99999999
1 cos x
x
=1
5
55.
-5
5
-5
(a)
lim f (x) =
x→−1+
(b) lim f (x)
∞
≈ 2.7
x→0
(c) lim f (x) = 1 x→∞
1 π π 56. (a) The area of the right triangle shown in Figure 2.5.18 is r2 sin cos . Since there are 2 n n 2n such right triangles, the area of the polygon is: A(n) = 2n 2
(b) A(100)
≈ 3.1395r ;
1 2 π π r sin cos = nr 2 2 n n
A(1000)
≈ 3.1416r
1 2π n 2π sin = r 2 sin 2 n 2 n
2
(c) Letting x = 2π/n (while noting that n = 2π/x) and substituting into A(n) above, we obtain: π sin x A(n) = r2 sin x = πr 2 x x
From (10) of Section 2.4, we see that: lim A(n) = πr
2
n→∞
sin x lim = πr 2 x→0 x
x2 57. (a) lim [f (x) g(x)] = lim (x 1) x→±∞ x→±∞ x + 1 x2 (x 1)(x + 1) x2 (x2 1) = lim = lim x→±∞ x + 1 x→±∞ x + 1 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.
−
− −
(c) g is a slant asymptote to f .
− −
−
−
| |
98
CHAPTER 2. LIMIT OF A FUNCTION
58. All points P are of the form (x, x2 + 1) while all points Q are of the form (x, x2 ). When the y coordinates of P and Q are the same, we have x2P + 1 = x2Q or xQ = x2P + 1, and thus the horizontal distance between P and Q is xQ xP = x2P + 1 xP . Thus: lim
x→∞
|
x2 + 1
→∞
x→∞
2.6
2
= lim
| − | | − | √ x + 1 + x √ x + 1 + x x +1−x x +1−x 1 √ = lim √ x x + 1 + x x + 1 + x 2
− x| = xlim
2
2
Limits — A Formal Approach
2 2
→∞
2
= 0.
1. 10
| − 10| = 0 < for any choice of δ . 2. |π − π | = 0 < for any choice of δ . 3. |x − 3| < whenever 0 < |x − 3| < . Choose δ = . 4. |2x − 8| = 2|x − 4| < whenever 0 < |x − 4| < /2. Choose δ = /2. 5. |x + 6 − 5| = |x + 1| < whenever 0 < |x − (−1)| < . Choose δ = . 6. |x − 4 − (−4)| = |x − 0| < whenever 0 < |x − 0| < . Choose δ = . 7. |3x + 7 − 7| = 3|x − 0| < whenever 0 < |x − 0| < /3. Choose δ = /3. 8. |9 − 6x − 3| = |6 − 6x| = 6|x − 1| < whenever 0 < |x − 1| < /6. Choose δ = /6. 2x − 3 1 1 1 − 9. = |2x − 4| = |x − 2| < whenever 0 < |x − 2| < 2. Choose δ = 2. 4 4 4 2
10. 8(2x + 5)
|
−
− 48| = |16x − 8| = 16 x
x2 25 11. x + 5
− − − | − | − |
1 < whenever 0 < x 2
| − 2| < /16. Chose δ = /16.
− − (−10) = |x − 5+10| = |x − (−5)| < whenever 0 < |x − (−5)| < . Choose δ = .
x2
7x + 12 1 (x 3)(x 4) 1 1 12. = + = x 2x 6 2 2(x 3) 2 2 0 < x 3 < 2. Choose δ = 2. 13.
8x5 + 12x4 x4
−
−
−
| − 4 + 1| = 12 |x − 3 | < whenever
12 = 8x + 12 12 = 8 x 0 < whenever 0 < x 0 < /8. Choose δ = /8.
2x3 + 5x2 2x 14. x2 1 whenever 0 < x
− |
|−| 2
| − |
− − 5 − 7 = (2x + 5)(x − 1) − 7 = |2x + 5 − 7| = |2x − 2| = 2|x − 1| < − x −1 | − 1| < /2. Choose δ = /2. √ √ 15. |x − 0| = |x − 0| < whenever 0 < |x − 0| < . Choose δ = . √ √ 16. |8x − 0| = 8|x − 0| < whenever 0 < |x − 0| < /2. Choose δ = /2. 2
2
3
3
2
3
3
99
2.6. LIMITS — A FORMAL APPROACH
|√ 5x − 0| = √ 5|x − 0| / < whenever 0 < x < /5. Choose δ = /5. √ √ 18. | 2x − 1 − 0| = 2|x − 1/2| / < whenever 1/2 < x < 1/2 + /2. Choose δ = /2. 19. |2x − 1 − (−1)| = |2x| = 2|x − 0| < whenever 0 − /2 < x < 0. Choose δ = /2. 20. |3 − 3| = 0 < whenever x > 1, for any choice of δ . 21. Note that |x − 9| = |x − 3||x + 3| and consider only values of x for which |x − 3| < 1. Then 2 < x < 4 and 5 < x + 3 < 7, so |x + 3 | < 7. Thus, |x − 9| = |x − 3||x + 3 | < 7 |x − 3| < whenever |x − 3| < /7. Choose δ = min {1, /7}. 22. Note that |2x + 4 − 12| = 2|x − 4| = 2|x − 2||x + 2| and consider only values of x for which |x − 2| < 1. Then 1 < x < 3 and 3 < x + 2 < 5, so |x + 2| < 5. Thus |2x + 4 − 12| = 2|x − 2||x + 2| < 10 |x − 2| < whenever |x − 2| < /10. Choose δ = min{1, /10}. √ √ 23. Note that |x − 2x + 4 − 3| = |x − 1| < whenever |x − 1| < . Choose δ = . 24. Note that |x +2x − 35| = |x − 5||x+7 | and consider only values of x for which |x − 5| < 1. Then 4 < x < 6 and 11 < x +7 < 13, so |x +7| < 13. Thus |x +2x − 35| = |x − 5||x +7| < 13 |x − 5| whenever |x − 5| < /13. Choose δ = min {1, /13}. √ √ 25. We √ need to show | x − a| < whenever 0 < |x − a| < δ for an appropriate choice of δ . For δ = a, we have √ x + √ a |x − a| |x − a| √ a √ √ √ √ | x − a| = | x − a| · √ √ = √ √ < √ < √ = 1 2
17.
2
2
1 2
2
2
2
2
2
2
2
2
2
2
2
x +
a
x +
a
a
a
| − a| < δ . Thus, xlima √ x = √ a. 26. We need to show that |1/x − 1/2| < , whenever 0 < |x − 2| < δ , for an appropriate choice of δ . Without loss of generality, we may assume that δ < 1. Then |x − 2| < 1 or 1 < x < 3. For whenever 0 < x
→
these values of x, 1/3 < 1/x < 1. Then, for δ = 2, we have
− 1 x
whenever 0 < x
| − |
1 1 = 2 2
1 x
2
x <
1 (1) x 2
| − 2| < 12 (2) =
| − 2| < δ . Thus, xlim 1/x = 1/2. →2
27. Assume lim f (x) = L. Take = 1. Then there exists δ > 0 such that f (x) 0 < x
|
x→1
− L| < 1 whenever
| − 1| < δ . To the right of 1, choose x = 1 + δ/2. Since 0 < |1 + δ/2 − 1| = |δ/2| < δ, |f (1 + δ/2) − L| = |0 − L| = |L| < 1, we must have −1 < L < 1. or To the left of 1, choose x = 1 − δ/2. Since 0 < |1 − δ/2 − 1| = | − δ/2| < δ, |f (1 − δ/2) − L| = |2 − L| < 1, we must have or
1 < L < 3.
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100
CHAPTER 2. LIMIT OF A FUNCTION
Since no L can satisfy the conditions that lim f (x) does not exist.
−1 < L < 1 and 1 < L < 3, we conclude that
x→1
28. Assume lim f (x) = L. Take = 1. Then there exists δ > 0 such that f (x) 0 < x
|
x→3
− L| < 1 whenever
| − 3| < δ . To the right of 3, choose x = 3 + δ/2. Since 0 < |3 + δ/2 − 3| = |δ/2| < δ, |f (3 + δ/2) − L| = |− 1 − L| = |L + 1| < 1, we must have −2 < L < 0. or To the left of 3, choose x = 3 − δ/2. Since 0 < |3 − δ/2 − 3| = |− δ/2| < δ, |f (3 − δ/2) − L| = |1 − L| = |L − 1| < 1, we must have or
0 < L < 2.
Since no L can satisfy the conditions that lim f (x) does not exist.
−2 < L < 0 and 0 < L < 2, we conclude that
x→3
29. Assume lim f (x) = L. Take = 1. Then there exists δ > 0 such that f (x) 0 < x
|
x→0
− L| < 1 whenever
| − 0| < δ . To the right of 0, choose x = δ/2. Since 0 < |δ/2 − 0| = |δ/2| < δ, |f (δ/2) − L| = |2 − δ/2 − L| < 1, we must have or 1 − δ/2 < L < 3 − δ/2. To the left of 0, choose x = −δ/2. Since 0 < |− δ/2 − 0| = |− δ/2| < δ, |f (−δ/2) − L| = |− δ/2 − L| < 1, we must have −1 − δ/2 < L < 1 − δ/2. or Since no L can satisfy the conditions that 1 − δ/2 < L < 3 − δ/2 and − 1 − δ/2 < L < 1 − δ/2, we conclude that lim f (x) does not exist. x→0
30. Assume lim f (x) = L. Take = 1. Then there exists δ > 0 such that f (x) x→0
|
− L| < 1 whenever
0 < x 0 < δ . Since f (x) L < 1 for all x such that 0 < x < δ , we may assume that δ < 2. To the right of 0, choose x = δ/2.
| − |
|
− |
Since we must have or To the left of 0, choose x = Since we must have or
| |
0 < δ/2 0 = δ/2 < δ, L = 2/δ L = L 2δ < 1, 2/δ 1 < L < δ/2 + 1.
| − | | | |f (δ/2) − | | − | | − | − −δ/2. 0 < L =
|− δ/2 − 0| = |δ/2| < δ, |f (−δ/2) − | |− 2/δ − L| = |L + 2/δ | < 1, −2/δ − 1 < L < −2/δ + 1.
101
2.6. LIMITS — A FORMAL APPROACH
Since we assumed δ < 2, we have 1 < 2/δ 1 > 2/δ
and
or or
− −
0 < 2/δ 1 0 > 2/δ + 1.
−
−
Having established 2/δ
− 1 < L < δ/2 + 1 and −2/δ − 1 < L < −2/δ + 1, these imply −2/δ − 1 < L < 0. 0 < L < 2δ + 1 and
Since it is impossible for L to satisfy both of these inequalities, lim f (x) does not exist. x→0
31. By Definition 2.6.5(i), for any > 0 we must find an N > 0 such that
Now by considering x > 0,
5x 1 2x + 1
− − 5 2
5x 1 2x + 1
− − 5 2
< whenever x > N.
−
7 7 7 = = < < 4x + 2 4x + 2 4x
whenever x > 7/4. Hence, choose N = 7/4.
32. By Definition 2.6.5(i), for any > 0 we must find an N > 0 such that
Now by considering x > 0,
2x 3x + 8
2x 3x + 8
−
−
2 < whenever x < N. 3
−
2 16 16 16 = = < < 3 9x + 24 9x + 24 9x
whenever x > 16/9. Hence, choose N = 16/9.
33. By Definition 2.6.5(ii), for any > 0 we must find an N < 0 such that
Now by considering x < 0,
− − 10x x 3
whenever x <
10x x 3
− − 10
10 =
− 30
x
3
=
< whenever x < N.
−−
30 = ( x + 3)
−30/. Hence, choose N = −30/.
−
30 30 < < x + 3 x
34. By Definition 2.6.5(ii), for any > 0 we must find an N < 0 such that x2 x2 + 3
− − − −
Now by considering x < 0,
x2 x2 + 3
whenever x 2 > 3/ or x <
1 < whenever x < N.
3 3 3 1 = 2 = 2 < 2 < x +3 x +3 x
3/. Hence, choose N =
−
3/.
102
CHAPTER 2. LIMIT OF A FUNCTION
35. We need to show f (x) choice of δ . For δ = ,
|
− 0| = |f (x)| < whenever 0 < |x − 0| = |x| < δ for an appropriate
|f (x)| =
| |
x , x rational < whenever 0 < x < δ. 0, x irrational
| |
Thus, lim f (x) = 0. x→0
2.7
The Tangent Line Problem 10
1.
10
-5
5
10
-10
change in x = h = 2.5 2 = 0.5 change in y = f (2 + 0.5) f (2) = 2.75 5 = change in y 2.25 msec = = = 4.5 change in x 0.5
−
−
−
− −
−2.25
10
2.
10
-5
5
10
-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 msec = = = change in x 1/4 4
−−
−
−
10
3.
10
-5
5
10
-10
change in x = h = 1 ( 2) = 1 change in y = f ( 2 + 1) f ( 2) = change in y 7 msec = = =7 change in x 1
− − − − − −
−1 − (−8) = 7
5
4.
-5
5
-5
change in x = h = 1 0.9 = 0.1 change in y = f (1 + 0.1) f (1) = 10/11 1 = change in y 1/11 10 msec = = = change in x 1/10 11
−
−
−
− −
−1/11
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103
2.7. THE TANGENT LINE PROBLEM
3
5.
!
2π π π = 3 2 6 2 π π π change in y = f + f = sin π 1 = 3/2 2 6 2 3 change in y 3/2 1 3 3 6 msec = = = change in x π/6 π
change in x = h =
!
2
−
−
-3
− √ −
√ −
√
−1
3
6.
–!
π π π change in x = h = = 3 2 6 π π π π 1 change in y = f + f = cos 3 6 3 6 2 3 1 3 1 = = 2 2 2 change in y ( 3 1)/2 3 3 3 msec = = = change in x π/6 π
− − − − − − − −
–! 3
-3
√
7.
√ −
−
√ −
√ −
f (a) = f (3) = 3; f (a + h) = f (3 + h) = (h + 3) 2 f (a + h)
2
2
−6 + 6h + 9) − 6] − 3 = h
2
− f (a) = [(h + 3) − 6] − 3 = [(h + 6h = h(h + 6) f (a + h) − f (a) h(h + 6) m = lim = lim = lim (h + 6) = 6 tan
h With point of tangency (3, 3), we have y h→0
h→0 h 3 = 6(x 3) or y = 6x
h→0
− − 15. 8. f (a) = f (−1) = 7; f (a + h) = f (−1 + h) = −3(h − 1) + 10 f (a + h) − f (a) = [−3(h − 1) + 10] − 7 = [(−3h + 6h − 3) + 10] − 7 = −3h + 6h = h(6 − 3h) f (a + h) − f (a) h(6 − 3h) m = lim = lim = lim (6 − 3h) = 6 h h h h h With point of tangency ( −1, 7), we have y − 7 = 6(x + 1) or y = 6x + 13. 9. f (a) = f (1) = −2; f (a + h) = f (1 + h) = (h + 1) − 3(h + 1) f (a + h) − f (a) = [(h + 1) − 3(h + 1)] − (−2) = (h − h − 2) − (−2) = h − h = h(h − 1) f (a + h) − f (a) h(h − 1) m = lim = lim = lim (h − 1) = −1 h h h h h With point of tangency (1, −2), we have y + 2 = −(x − 1) or y = −x − 1. 10. f (a) = f (−2) = −17; f (a + h) = f (−2 + h) = −(h − 2) + 5(h − 2) − 3 f (a + h) − f (a) = [−(h − 2) + 5(h − 2) − 3] − (−17) = ( −h + 9h − 17) − (−17) = −h + 9h = h(9 − h) f (a + h) − f (a) h(9 − h) m = lim = lim = lim (9 − h) = 9 h h h h h With point of tangency ( −2, −17), we have y + 17 = 9(x + 2) or y = 9x + 1. −
2
2
tan
→0
2
2
→0
→0
2
2
tan
2
→0
2
→0
→0
2
2
2
tan
→0
2
→0
→0
104 11.
CHAPTER 2. LIMIT OF A FUNCTION 3
f (a) = f (2) =
−14; f (a + h) = f (2 + h) = −2(h + 2) + (h + 2) f (a + h) − f (a) = [−2(h + 2) + (h + 2)] − (−14) = ( −2h − 12h − 23h − 14) − (−14) = h(−2h − 12h − 23) f (a + h) − f (a) h(−2h − 12h − 23) m = lim = lim 3
3
2
2
2
tan
h 12h
h→0
2
= lim ( 2h h→0
−
−
h
h→0
− 23) = −23
With point of tangency (2, 14), we have y + 14 =
−23(x − 2) or y = −23x + 32. f (a) = f (1/2) = −3; f (a + h) = f (1/2 + h) = 8(h + 1/2) − 4 f (a + h) − f (a) = [8(h + 1/2) − 4] − (−3) = (8h + 12h + 6h − 3) − (−3) = 2h(4h + 6h + 3) f (a + h) − f (a) 2h(4h + 6h + 3) m = lim = lim −
12.
3
3
3
2
2
2
tan
h→0 h = lim 2(4h + 6h + 3) = 6
h
h→0
2
h→0
With point of tangency (1/2, 3), we have y + 3 = 6(x
−
13.
− 1/2) or y = 6x − 6.
−1/2; f (a + h) = f (−1 + h) = 2(h 1− 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 · m = lim = lim h h h 2(h − 1) h f (a) = f ( 1) =
−
tan
→0
= lim
→0
1
=
− 1) 1 1 x With point of tangency ( −1, −1/2), we have y + = − (x + 1) or y = − . 2 2 2 h→0
14.
2(h
− 12
4 (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→0 h + 1 h h→0 h + 1 h With point of tangency (2, 4), we have y 4 = 4(x 2) or y = 4x + 12. f (a) = f (2) = 4; f (a + h) = f (2 + h) =
−
− − −
− − −
15.
− −
f (a) = f (0) = 1; f (a + h) = f (h) =
− ·
1
−
− −
−
(h 1)2 1 h2 + 2h h(2 h) f (a + h) f (a) = 1 = = 2 2 (h 1) (h 1) (h 1)2 f (a + h) f (a) h(2 h) 1 2 mtan = lim = lim = lim h→0 h→0 (h h→0 (h h 1)2 h With point of tangency (0, 1), we have y 1 = 2(x 0) or y = 2x + 1.
−
− −
−
−
−
−
−
− − − · − −
−
−h − 1)
2
=2
105
2.7. THE TANGENT LINE PROBLEM
− −18+ h −8 − 8 = −8 − 8h + 8 = −8h 8 − f (a + h) − f (a) = 4 − 12 = −1 + h −1 + h h−1 h−1 −8h · 1 = lim −8 = 8 f (a + h) − f (a) m = lim = lim h h h h h−1 h h−1 With point of tangency ( −1, 12), we have y − 12 = 8(x + 1) or y = 8x + 20. √ 17. f (a) = f (4) = 2; f (a + h) = f (4 + h) = 4 + h √ 4 + h + 2 4 + h − 4 √ √ f (a + h) − f (a) = 4 + h − 2 = ( 4 + h − 2) √ = √ = √ 16.
f (a) = f ( 1) = 12; f (a + h) = f ( 1 + h) = 4
−
−
tan
→0
→0
→0
h 4 + h + 2
4 + h + 2 4 + h + 2 f (a + h) f (a) h 1 mtan = lim = lim h→0 h→0 h 4 + h + 2 h 1 1 = lim = h→0 4 4 + h + 2
√
−
·
√
With point of tangency (4, 2), we have y 18.
− 2 = 14 (x − 4) or y = 14 x + 1.
1 √ h + 1 √ √ √ 1 − h + 1 1 − h + 1 1 + h + 1 − 1 = √ · √ = √
f (a) = f (1) = 1; f (a + h) = f (1 + h) = f (a + h)
1 − f (a) = √ h + 1 1−h−1 = √
h + 1
=
h + 1
h + 1
h √ h + 1−+ h + 1
h + 1 + h + 1 f (a + h) f (a) mtan = lim = lim h→0 h→0 h 1 1 = lim = h→0 2 h + 1 + h + 1
−
√ −
√
−h
1 h + 1 + h + 1 h
·
−
With point of tangency (1, 1), we have y 19.
1+
− 1 = −12 (x − 1) or y = −12 x + 32 .
f (a) = f (π/6) = 1/2; 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→0 h 2 h 2 h
−
− √
− √
−
= (1/2)(0) + ( 3/2)(1) =
− √
−
√
−
·
√
·
3/2
With point of tangency (π/6, 1/2), we have y
−
√
1 3 = x 2 2
√
π 3 or y = x 6 2
−
−
√ 3π
1 + . 12 2
www.elsolucionario.org
106 20.
CHAPTER 2. LIMIT OF A FUNCTION
f (a) = f (π/4) = f (a + h)
−
√
2/2; f (a + h) = f (π/4 + h) = cos(π/4 + h)
− √
√
π 2 π π 2 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→0 2 h h h
√
√
−
−
√ √
−
√
√
−
−
− − − −
√ − 1) = − 2/2 √ 2 √ 2 π √ With point of tangency (π/4, 2/2), we have y − = − x− 2 2 4 √ 2 = ( 2/2)(0
2 21.
.
or y =
√ 2
− 2
x+
√ 2π 8
+
f (a) = f (1) = 1; f (a + h) = f (1 + h) = (h + 1)2 f (a + h)
2
2
− f (a) = [(h + 1) ] − 1 = (h + 2h + 1) − 1 = h(h + 2) f (a + h) − f (a) h(h + 2) m = lim = lim = lim (h + 2) = 2 tan
h→0 h→0 h h The slope of the tangent at the blue point (1 , 1) is 2. The slope of the line through (1, 1) and (4, 6) is m = (6 1)/(4 1) = 5/3. Since the slopes are not equal, then this line is not tangent to the graph. h→0
−
−
22. Since there is more than one line, we first find the slope of the tangent line at the “general point” (a, f (a)). f (a) = a 2 ; f (a + h) = (h + a)2 f (a + h)
2
2
2
2
2
− f (a) = [(h + a) ] − (a ) = (h + 2ha + a ) − a = h(h + 2a) f (a + h) − f (a) h(h + 2a) m = lim = lim = lim (h + 2a) = 2a tan
h→0 h→0 h h Now that we have determined that mtan = 2a, then the slope of the tangent at the blue point ( 1, 1) is m tan ( 1) = 2( 1) = 2. The slope of the line through ( 1, 1) and (1, 3) is m = ( 3 1)/(1 + 1) = 2. Since the slopes are equal, then this line is tangent to the graph.
− −−
h→0
−
−
−
−
−
−
The slope of the tangent at the blue point (3 , 9) is m tan (3) = 2(3) = 6. The slope of the line through (3, 9) and (1, 3) is m = (9 + 3)/(3 1) = 6. Since the slopes are equal, then this line is tangent to the graph.
−
−
23. We know that the points (2, 0) and (6, 4) are on the tangent line, so its equation is
− 0 = 02 −− 46 (x − 2) The line’s y-intercept is (0, −2). y
or
y = x
−2
24. We know that the points (0, 4) and (7, 0) are on the tangent line, so its equation is y
− 0 = 40 −− 07 (s − 7)
or
y =
−47 x + 4
107
2.7. THE TANGENT LINE PROBLEM
Since the point of tangency ( 5, f ( 5)) is on this tangent line, then
−
−
f ( 5) =
−
25.
−47 (−5) + 4 = 487
2
f (a) =
2
−a + 6a + 1; f (a + h) = −(h + a) + 6(h + a) + 1 f (a + h) − f (a) = [−(h + a) + 6(h + a) + 1] − (−a + 6a + 1) = −h − 2ha − a + 6h + 6a + 1 − (−a ) − 6a − 1 = −h − 2ha + 6h = h(−h − 2a + 6) f (a + h) − f (a) h(−h − 2a + 6) = lim m = lim 2
2
2
2
2
2
tan
h→0
= lim (−h − h →0
h 2a + 6) =
h
h→0
−2a + 6
The tangent line is horizontal when m tan = 0, so we substitute and solve m tan = 0 = 2a + 6, yielding 2a = 6 and a = 3. Thus, the tangent line is horizontal at (3 , f (3)) = (3, 10).
−
26.
f (a) = 2a2 + 24a
2
− 22; f (a + h) = 2(h + a) + 24(h + a) − 22 f (a + h) − f (a) = [2(h + a) + 24(h + a) − 22] − (2a + 24a − 22) = 2h + 4ha + 2a + 24h + 24a − 22 − 2a − 24a − (−22) 2
2
2
2
2
= 2h2 + 4ha + 24h = h(2h + 4a + 24)
f (a + h) f (a) h(2h + 4a + 24) = lim h→0 h→0 h h = lim (2h + 4a + 24) = 4a + 24
−
mtan = lim
h→0
The tangent line is horizontal when m tan = 0, so we substitute and solve m tan = 0 = 4a +24, yielding 4a = 24 and a = 6. Thus, the tangent line is horizontal at ( 6, f ( 6)) = ( 6, 94).
− −
27.
−
−
−
f (a) = a3
−
3
− 3a; f (a + h) = (h + a) − 3(h + a) f (a + h) − f (a) = [(h + a) − 3(h + a)] − (a − 3a) = h + 3h a + 3ha + a − 3h − 3a − a − (−3a) = h + 3h a + 3ha − 3h = h(h + 3ah + 3a − 3) f (a + h) − f (a) h(h + 3ah + 3a − 3) m = lim = lim 3
3
3
2
2
3
2
2
3
3
2
2
2
tan
h = lim (h + 3ah + 3a2 h→0
2
h→0
2
h
h→0
2
− 3) = 3a − 3
The tangent line is horizontal when m tan = 0, so we substitute and solve m tan = 0 = 3a2 3, yielding 3a2 = 3 and a = 1. Thus, the tangent line is horizontal at ( 1, f ( 1)) = ( 1, 2) and (1, f (1)) = (1, 2).
−
±
−
−
−
−
108 28.
CHAPTER 2. LIMIT OF A FUNCTION 3
f (a) =
2
3
2
−a + a ; f (a + h) = −(h + a) + (h + a) f (a + h) − f (a) = [−(h + a) + (h + a) ] − (−a + a ) = −h − 3h a − 3ha − a + h + 2ah + a − (−a ) − a = −h − 3h a − 3ha + h + 2ah = h(−h − 3ah − 3a + h + 2a) f (a + h) − f (a) h(−h − 3ah − 3a + h + 2a) m = lim = lim 3
2
3
3
2
2
3
3
2
2
2
2
2
2
3
2
2
2
tan
h→0
2
= lim ( h h→0
− −
h 3ah
2
h 3a + 2a
h→0
2
− 3a
+ h + 2a) =
2
2
−
The tangent line is horizontal when mtan = 0, so we substitute and solve mtan = 0 = 3a2 +2a, yielding a(3a 2) = 0 and a = 0, 2/3. Thus, the tangent line is horizontal at (0 , f (0)) = (0, 0) and (2/3, f (2/3)) = (2/3, 4/27).
−
−
29. vave =
change of distance 290 mi = = 58 mi/h change in time 5h
30. vave =
change of distance 1/2 mi (1/2 mi) 1/2 mi = = = = 45 mi/h change in time 40 s (40 s)/(3600 s/h) 1/90 h
The car will not be stopped for speeding. 31. vave =
change of distance 3500 km ; 920 km/h = ; t change in time t
32. vave =
change of distance 20 mi = 1 change in time 36 h
33. ∆s = s(t0
≈ 3.8 h = 3 h 48 min
− 10 mi = 20 mi − 10 mi = 10 mi = 6 mi/h − 1 h 19/6 h − 3/2 h 5/3 h +∆t) − s(t ) = f (3+∆t) − f (3) = [−4(3+∆t) +10(3+∆t)+6] − 0 = −14∆t − 4∆t 1 2
2
0
2
The instantaneous velocity at t = 3 is ∆s v(3) = lim = lim ∆t→0 ∆t ∆t→0
−14∆t − 4∆t ∆t
2
= lim ( 14 ∆t→0
1 34. ∆s = s(t0 + ∆t) s(t0 ) = f (∆t) f (0) = ∆t + 5∆t + 1 The instantaneous velocity at t = 0 is
−
2
−
−
− − 4∆t) = −14.
5∆t3 + ∆t2 5∆t 1= = 5∆t + 1
−
−5
∆s 5∆t3 + ∆t2 5∆t 5∆t2 + ∆t 5 = lim = lim = ∆t→0 ∆t ∆t→0 ∆t→0 (5∆t + 1)∆t 5∆t + 1
−
v(0) = lim
35. (a) ∆s = s(t0 + ∆t)
−
− s(t ) = f (1/2 + ∆t) − f (1/2) = −4.9(1/2 + ∆t) − 4.9∆t 0
= 4.9∆t2 The instantaneous velocity at t = 1/2 is
−
∆s = lim ∆t→0 ∆t ∆t→0
v(1/2) = lim
2
−5.
+ 122.5
− 121.275
2
−4.9∆t − 4.9∆t = ∆t
lim ( 4.9∆t
∆t→0
−
− 4.9) = −4.9 m/s.
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109
2.7. THE TANGENT LINE PROBLEM
(b) The ball hits the ground when s(t) = 0: 2
−4.9t
+ 122.5 = 0; t2 = 122.5/4.9; t = 5 s.
(c) Since the ball impacts at t = 5, ∆s = s(t0 + ∆t) 2
=
2
− s(t ) = f (5 + ∆t) − f (5) = [−4.9(5 + ∆t) 0
+ 122.5]
−49∆t − 49∆t
2
− [−4.9(5)
+ 122.5]
The impact velocity at t = 5 is ∆s = lim v(5) = lim ∆t→0 ∆t ∆t→0 36. (a) Setting
−
1 gt2 2
∆t
lim ( 49∆t
∆t→0
−
+ h = 0 and solving for t > 0, we obtain t =
(b) Earth:
timpact =
Mars:
timpact =
2(100)/32 = 2.5 s
≈ 4.08 s Moon: t = 2(100)/5.5 ≈ 6.03 s 1 ∆s = s(t + ∆t) − s(t ) = − g(t + ∆t) 2 impact
(c)
2
−49∆t − 49∆t =
0
− 49) = −49 m/s.
2h/g.
2(100)/12
0
0
2
+ h
− (− 12 gt + h) = − 12 g∆t − gt ∆t 2 0
2
0
The instantaneous velocity at t impact is ∆s v(timpact ) = lim = lim ∆t→0 ∆t ∆t→0
− 12 g∆t − gt ∆t 2
∆t
0
= lim
∆t→0
−
(d) The impact velocities are vEarth = vMars vMoon
−(32)(2.5) = −80 ft/s ≈ −(12)(4.08) = −48.96 ft/s ≈ −(5.5)(6.03) = −33.165 ft/s.
37. (a)
s(t) =
2
−16t + 256t s(2) = −16(2 ) + 256(2) = 448 ft s(6) = −16(6 ) + 256(6) = 960 ft s(9) = −16(9 ) + 256(9) = 1008 ft s(10) = −16(10 ) + 256(10) = 960 ft s(5) = −16(5 ) + 256(5) = 880 ft 2 2 2
2
(b)
2
s(2) = 448 ft [from (a)] change of distance 880 ft vave = = change in time 5s
− 448 ft = 144 ft/s −2 s
1 g∆t 2
− gt
0
=
−gt . 0
110
CHAPTER 2. 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 9s 7s 2 At t = 7 s, the projectile is at a height of 1008 ft on its way upward. After it reaches a maximum height, it begins to fall downward and, at t = 9 s, the height is once again 1008 ft. Since distance upward is positive and distance downward is negative, the net distance is zero. (d) The projectile hits the ground when s(t) = 0:
−
− −
2
−16t
+ 256t = 0; 16t2 = 256t; t = 256/16 = 16 s
(e) For some general time t: ∆s = s(t + ∆t) s(t) = [ 16(t + ∆t)2 + 256(t + ∆t)] 2
−
−
= 16∆t + 256∆t 32t∆t = ∆t( 16∆t + 256 The instantaneous velocity at a general time t is
−
−
−
− (−16t
2
+ 256t)
− 32t)
∆s ∆t( 16∆t + 256 32t) = lim ∆t→0 ∆t ∆t→0 ∆t = lim ( 16∆t + 256 32t) = (256 32t) ft/s.
−
v(t) = lim
∆t→0
−
−
−
−
(f) From (d), the pro jectile impacts at t = 16 s. From (e), v(t) = 256 32t so v(16) = 256 32(16) = 256 ft/s. (g) The maximum height is reached when v(t) = 0: 256 32t = 0 gives us t = 8 s. Since s(t) = 16t2 + 256t, we have s(8) = 16(82 ) + 256(8) = 1024 ft.
−
38.
−
−
−
− − (a) s(4) ≈ 1.3 ft; s(6) ≈ 2.7 ft s(6) − s(4) 2.7 − 1.3 (b) v ≈ = = 0.7 ft/s 6−4 2 ave
(c) The instantaneous velocity at t = 0 is the slope of the tangent line to the graph at t = 0. In this case, v 0 1 ft/s. (d) t 3 s (e) The velocity is decreasing where the slopes of the tangent lines are decreasing; in this case, for 0 < t < 3. (f) The velocity is increasing where the slopes of the tangent lines are increasing; in this case, for 3 < t < 7.
≈
≈
39. The slopes m of a tangent line at ( a, f (a)) and m of a tangent line at ( a, f ( a)) are:
− − f (−a + h ) − f (−a)
f (a + h) f (a) ; m = lim h→0 h →0 h h As defined in Section 1.2, an even function is a function which is symmetric with respect to the y-axis: f ( x) = f (x) for all x. Since f is even, then f ( a) = f (a) and f ( a + h ) = f ( [ a + h ]) = f (a h ), resulting in:
−
m = lim
−−
−
−
−
m = lim h
→0
f (a
− h ) − f (a) =
h
f (a + [ h ]) lim h →0 h
−
−
− f (a)
111
CHAPTER 2 IN REVIEW
Without loss of generality, we apply the substitution h =
−h to obtain: f (a + [ −h ]) − f (a) f (a + h) − f (a) m = lim = lim = −m h h −h h 40. The slopes m of a tangent line at ( a, f (a)) and m of a tangent line at ( −a, f (−a)) are: f (a + h) − f (a) f (−a + h ) − f (−a) m = lim ; m = lim
→0
→0
h
h→0
h
h
→0
As defined in Section 1.2, an odd function is a function which is symmetric with respect to the origin: f ( x) = f (x) for all x. Since f is odd, then f ( a) = f (a) and f ( a + h ) = f ( [ a + h ]) = f (a h ), resulting in:
−
− − − − − −f (a − h ) − [−f (a)] = − lim −[f (a − h ) + f (a)] m = lim
− −−
−
h
h
→0
h
→0
h
Without loss of generality, we apply the substitution h =
−h to obtain: −[f (a − h ) + f (a)] = lim −[f (a + h) − f (a)] = m m = lim h h −h h 41. To show that the graph of f (x) = x + |x| does not possess a tangent line at (0 , 0), we examine f (0 + h) − f (0) [(0 + h) + |0 + h|] − 0 h + |h| lim = lim = lim
→0
→0
2
2
h→0
h
2
h
h→0
h
h→0
From the definition of absolute value, we see that lim
h→0+
h2 + h h2 + h = = h + 1 = 1 h h whereas
||
h2 + h h2 h lim = = h h h h→0 −
||
−
− 1 = −1 f (0 + h) h→0 h
Since the right-hand and left-hand limits are not equal, we conclude that lim
h2 + h lim does not exist, and that therefore f has no tangent line at (0 , 0). h→0 h
||
Chapter 2 in Review A. True/False 1. True 2. False; lim
√ x − 5 = 0.
3. False; lim
|x| = −1.
x→5+
x→0
−
x
− f (0) =
www.elsolucionario.org
112
CHAPTER 2. LIMIT OF A FUNCTION 2
4. False; lim e2x−x = 0. x→∞
5. False; lim
x→0+
−1
tan
1 π = . x 2
6. True 7. True 8. False; let f (x) = 0. 9. False; consider f (x) = 10. False; consider f (x) = 11. False; consider f (x) =
1 1 , g(x) = 4 , and a = 0. 2 x x 1 1 , g(x) = , and a = 0. 2 x tan2 x
−x.
12. True 13. True; since f ( 1) < 0 and f (1) > 0.
−
14. False; consider f (x) = 1 and g(x) = x
− 2.
15. True 16. False; consider f (x) =
17. False; consider f (x) = 18. True; since lim [(x x→a
−
1, x < 0 and a = 0. 1, x > 0
1, x 3 . 2, x > 3
≤
− a)f (x)] = [lim (x − a)][lim f (x)] = 0 · f (a) = 0. x a x a →
→
19. True 20. False; lim f (x) = 4 = f (5). x→5
21. False; since
√ x x + 1
is undefined for x < 0.
22. False; the slope m of the tangent line at (3 , f (3)) is 1. There is not enough information to determine the value of f (3).
113
CHAPTER 2 IN REVIEW
B. Fill in the Blanks 1. 4 2. 1 3. -1/5 4. -1/2 5. 0 6. 3/5 7.
∞
8. 0 9. 1 10. 1/4 11. 3− 12. 4 13.
−∞
14. 0+ 15.
−2
16. Dividing by x 2 we have 1
x2 3
− ≤
f (x) x2
f (x) Theorem we have lim 2 = 1. x→0 x
≤ 1. Since xlim
→0
x 2 = 1 = lim 1, by the Squeeze x→0 3
− 1
17. 10 18. 8 19. continuous 20. 2 21. 9 22. Since f (x) = x 2 is continuous, lim f (g(x)) = f ( lim g(x)) = f ( 9) = ( 9)2 = 81. x→−5
x→−5
−
−
114
CHAPTER 2. LIMIT OF A FUNCTION
C. Exercises 5 5
1.
2. -5
-5
5
5
-5 -5
3 5
3.
4.
-3
3
-5
-3
5. (a), (e), (f), (h) 6. (b), (e), (h) 7. (c), (h) 8. (b), (c), (d), (e), (f), (i) 9. (b), (c), (d), (e), (f) 10. (a), (g), (j) 5
11.
-5
5
The function is continuous everywhere.
-5
6
12.
The function is discontinuous at x = 2 and x = 4.
3
3
13. (
6
−∞, −1), (−1, 0), (0, 1), and (1, ∞)
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115
CHAPTER 2 IN REVIEW
14. [ 2, 1) and (1, 2]
− √ √ 15. (−∞, − 5) and ( 5, ∞) 16. (nπ,nπ + π) for n = 0, 1, 2, . . . 17. For f (x) to be continuous at the number 3, we must have f (3) = 3k + 1 = lim (2 + x→3
Thus, we must solve for k in the equation 3k + 1 = 2
− kx).
− 3k, resulting in k = 1/6. Therefore: x ≤ 3
x + 1, 6 x 2 , x > 3 6
−
f (x) =
18. For f (x) to be continuous everywhere, we must have f (1) = 5 = lim (ax + b) and f (3) = 3a + b = lim (3x x→3+
an b yields a =
x→1+
− 8). Thus, we get two equations 5 = a + b and 1 = 3a + b. Solving for a
−2, b = 7. Therefore:
f (x) =
19.
−
x + 4, x 1 2x + 7, 1 < x 3x 8, x > 3
≤
≤ 3
− f (a) = f (2) = 32; f (a + h) = f (2 + h) = −3(h + 2) + 16(h + 2) + 12 f (a + h) − f (a) = [−3(h + 2) + 16(h + 2) + 12] − 32 = −3h − 12h − 12 + 16h + 32 + 12 − 32 = −3h + 4h = h(−3h + 4) f (a + h) − f (a) h(−3h + 4) m = lim = lim =4 2
2
2
tan
2
h With point of tangency (2, 32), we have y 20.
h→0
h 32 = 4(x
h→0
− 2) or y = 4x + 24. f (a) = f (−1) = −2; f (a + h) = f (−1 + h) = (h − 1) − (h − 1) f (a + h) − f (a) = [(h − 1) − (h − 1) ] − (−2) = h − 3h + 3h − 1 − h + 2h − 1 − (−2) = h − 4h + 5h = h(h − 4h + 5) f (a + h) − f (a) h(h − 4h + 5) m = lim = lim =5 −
3
3
3
2
3
2
2
2
2
2
2
tan
h→0 h h With point of tangency ( 1, 2), we have y + 2 = 5(x + 1) or y = 5x + 3. h→0
− −
21.
1 −2; f (a + h) = f (1/2 + h) = 2(h +−1/2) −1 − (−2) = −1 − (−2) f (a + h) − f (a) = f (a) = f (1/2) =
2
2(h + 1/2)2 2h2 + 2h + 1/2 1 + 4h2 + 4h + 1 4h(h + 1) = = 2 2 2h + 2h + 1/2 2h + 2h + 1/2 f (a + h) f (a) 4h(h + 1) mtan = lim = lim =8 2 h→0 h→0 (2h + 2h + 1/2)h h
−
−
116
CHAPTER 2. LIMIT OF A FUNCTION
With point of tangency (1/2, 2), we have y + 2 = 8(x
− 1/2) or y = 8x − 6. √ 22. f (a) = f (4) = 12; 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) h − 16h + 64 − 16h − 64 √ √ = = (h − 8) − 4 h + 4 (h − 8) − 4 h + 4 h(h − 32) √ = (h − 8) − 4 h + 4 −32 = 2 f (a + h) − f (a) h(h − 32) √ m = lim = lim = h h −16 h [(h − 8) − 4 h + 4]h With point of tangency (4, 12), we have y − 12 = 2(x − 4) or y = 2x + 4. 23. f (a) = f (1) = 2; f (a + h) = f (1 + h) = −4(h + 1) + 6(h + 1) f (a + h) − f (a) = [−4(h + 1) + 6(h + 1)] − 2 = −4h − 8h − 4 + 6h + 6 − 2 = −4h − 2h = h(−4h − 2) f (a + h) − f (a) h(−4h − 2) m = lim = lim = −2 h h h h With point of tangency (1, 2), we have y − 2 = −2(x − 1) or y = −2x + 4. Thus, the line that is perpendicular to this line would have a slope of 1 /2 and also passes through (1, 2), resulting in the equation y − 2 = (x − 1)/2 or y = (x + 3)/2. 24. |2x + 5 − 7| = |2x − 2| = 2|x − 1| < whenever |x − 1| < /2. Thus, we choose δ = /2 and −
2
tan
2
→0
→0
2
2
2
2
tan
→0
→0
so δ = 0.005 when = 0.01. Finding δ proves that lim (2x + 5) = 7. x→1
