, Solution manual For Calculus Early
Transcendentals Single Variable 12th Edition
Howard Anton
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,Limits and Continuity
Exercise Set 1.1
1. (a) 2 (b) 2 (c) 2 (d) 2
2. (a) 1 (b) 1 (c) 1 (d) 1
3. (a) −1 (b) 3 (c) does not exist (d) 1
4. (a) 2 (b) 0 (c) does not exist (d) 2
5. (a) 4 (b) 4 (c) 4 (d) 2
6. (a) 1 (b) 1 (c) 1 (d) 0
7. (a) −∞ (b) −∞ (c) −∞ (d) 1
8. (a) +∞ (b) +∞ (c) +∞ (d) cannot be found from graph
9. (a) +∞ (b) +∞ (c) 2 (d) 2 (e) −∞ (f ) x = −2, x = 0, x = 2
10. (a) does not exist (b) −∞ (c) 0 (d) −1 (e) +∞ (f ) 3 (g) x = −2, x = 2
11. (i) −0.1 −0.01 −0.001 0.001 0.01 0.1
1.9866933 1.9998667 1.9999987 1.9999987 1.9998667 1.9866933
–0.1 0.1
2.01 2.01
1.99 1.99
(ii) –0.1 0.1 The limit appears to be 2.
1
,2 Chapter 1
12. (i) −0.5 −0.05 −0.005 0.005 0.05 0.5
−0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752
–0.5 0.5
–0.49 –0.49
–0.51 –0.51
(ii) –0.5 0.5 The limit appears to be −1/2.
13. (a) 2 1.5 1.1 1.01 1.001 0 0.5 0.9 0.99 0.999
0.1429 0.2105 0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.3337
0.9 1.1
0.37 0.37
0.333 0.333
0.3 0.3
0.9 1.1 The limit is 1/3.
(b) 2 1.5 1.1 1.01 1.001 1.0001
0.4286 1.0526 6.344 66.33 666.3 6666.3
1 1.1
100 100
0 0
1 1.1 The limit is +∞.
(c) 0 0.5 0.9 0.99 0.999 0.9999
−1 −1.7143 −7.0111 −67.001 −667.0 −6667.0
0.9 1
0 0
–100 –100
0.9 1 The limit is −∞.
,Exercise Set 1.1 3
14. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25
0.5359 0.5132 0.5001 0.5000 0.5000 0.4999 0.4881 0.4721
–0.1 0.1
0.51 0.51
0.5 0.5
0.49 0.49
–0.1 0.1 The limit is 1/2.
(b) 0.25 0.1 0.001 0.0001
8.4721 20.488 2000.5 20001
0 0.01
5000 5000
0 0
0 0.01 The limit is +∞.
(c) −0.25 −0.1 −0.001 −0.0001
−7.4641 −19.487 −1999.5 −20000
–0.01 0
0 0
–5000 –5000
–0.01 0 The limit is −∞.
15. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25
2.7266 2.9552 3.0000 3.0000 3.0000 3.0000 2.9552 2.7266
–0.05 0.05
3.01 3.01
2.99 2.99
–0.05 0.05 The limit is 3.
,4 Chapter 1
(b) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.001
1 1.7552 6.2161 54.87 541.1 −0.1415 −4.536 −53.19 −539.5
–1.01 –1 –0.99
2000 2000
–2000 –2000
–1.01 –1 –0.99 The limit does not exist.
16. (a) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.001
1.5574 1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.0000
–1.2 –1 –0.8
1.01 1.01
1 1
0.99 0.99
–1.2 –1 –0.8 The limit is 1.
(b) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25
1.9794 2.4132 2.5000 2.5000 2.5000 2.5000 2.4132 1.9794
–0.1 0.1
2.6 2.6
2.5 2.5
2.4 2.4
–0.1 0.1 The limit is 5/2.
17. False; define f (x) = x for x = a and f (a) = a + 1. Then limx→a f (x) = a = f (a) = a + 1.
18. True; by 1.1.3.
19. False; define f (x) = 0 for x < 0 and f (x) = x + 1 for x ≥ 0. Then the left and right limits exist but are unequal.
20. False; define f (x) = 1/x for x > 0 and f (0) = 2.
21. Answers may vary. One possible graph:
y
y = f(x)
x
,Exercise Set 1.1 5
22. Answers may vary. One possible graph:
y y = f(x)
2
1
x
–2 –1 1 2 3
23. Answers may vary. One possible graph:
y
y = f(x)
x
24. Answers may vary. One possible graph:
y
y = f(x)
x
25. Answers may vary. One possible graph:
y y = f(x)
2
x
–3 –1 1 3
–2
,6 Chapter 1
26. Answers may vary. One possible graph:
y
y = f(x)
x
x2 − 1
27. msec = = x − 1 which gets close to −2 as x gets close to −1, thus y − 1 = −2(x + 1) or y = 1 − 2(x + 1).
x+1
x2
28. msec = = x which gets close to 0 as x gets close to 0, thus y = 0.
x
x4 − 1
29. msec = = x3 + x2 + x + 1 which gets close to 4 as x gets close to 1, thus y − 1 = 4(x − 1) or y = 1 + 4(x − 1).
x−1
x4 − 1
30. msec = = x3 − x2 + x − 1 which gets close to −4 as x gets close to −1, thus y − 1 = −4(x + 1) or
x+1
y = 1 − 4(x + 1).
31. (a) The length of the rod while at rest.
(b) The limit is zero. The length of the rod approaches zero as its speed approaches c.
32. (a) The mass of the object while at rest.
(b) The limiting mass as the velocity approaches the speed of light; the mass is unbounded.
33. (a) As x → 4− , g(x) → 0+ , so lim− f (g(x)) = lim+ f (u) = 2
x→4 u→0
+ −
(b) As x → 4 , g(x) → 0 , so lim+ f (g(x)) = lim− f (u) = +∞
x→4 u→0
−
(c) As x → 0, g(x) → 2 , so lim f (g(x)) = lim f (u) = 2
x→0 u→2−
34. (a) As x → 4− , g(x) → 0+ , so lim f (g(x)) = lim f (u) = −1
x→4− u→0+
(b) As x → 4+ , g(x) → 0− , so lim f (g(x)) = lim f (u) = 0
x→4+ u→0−
−
(c) As x → 0, g(x) → 2 , so lim f (g(x)) = lim− f (u) = +∞
x→0 u→2
35. (a)
–0.001 0.001
3.5 3.5
3 3
2.5 2.5
–0.001 0.001
The limit appears to be 3.
,Exercise Set 1.2 7
(b)
–1.× 10–6 1.× 10–6
3.5 3.5
3 3
2.5 2.5
–1.× 10–6 1.× 10–6
The limit appears to not exist.
(c) Answers may vary with calculating utility. Using Mathematica:
x 0.001 (0.001)2 (0.001)3 (0.001)4 (0.001)5
f (x) 3.00416 3.00446 1.0 1.0 1.0
The limit appears to be 1.
(d) The limited precision of graphing and calculating utilities may compromise accurate calculation.
Exercise Set 1.2
1. (a) By Theorem 1.2.2, this limit is 2 + 2 · (−4) = −6.
(b) By Theorem 1.2.2, this limit is 0 − 3 · (−4) + 1 = 13.
(c) By Theorem 1.2.2, this limit is 2 · (−4) = −8.
(d) By Theorem 1.2.2, this limit is (−4)2 = 16.
√
3
(e) By Theorem 1.2.2, this limit is 6 + 2 = 2.
2 1
(f ) By Theorem 1.2.2, this limit is =− .
(−4) 2
2. (a) By Theorem 1.2.2, this limit is 0 + 0 = 0.
(b) The limit doesn’t exist because lim f doesn’t exist and lim g does.
(c) By Theorem 1.2.2, this limit is −2 + 2 = 0.
(d) By Theorem 1.2.2, this limit is 1 + 2 = 3.
(e) By Theorem 1.2.2, this limit is 0/(1 + 0) = 0.
(f ) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.
(g) The limit doesn’t exist because f (x) is not defined for 0 < x < 2.
√
(h) By Theorem 1.2.2, this limit is 1 = 1.
3. By Theorem 1.2.3, this limit is 2 · 1 · 3 = 6.
4. By Theorem 1.2.3, this limit is 23 + 2 · 2 − 5 = 7.
5. By Theorem 1.2.4, this limit is (32 − 2 · 3)/(3 + 1) = 3/4.
, 8 Chapter 1
6. By Theorem 1.2.4, this limit is (6 · 0 − 9)/(03 − 12 · 0 + 3) = −3.
x4 − 1
7. After simplification, = x2 + 1 (x = 1), and the limit is 12 + 1 = 2.
x2 − 1
t3 + 8
8. After simplification, = t2 − 2t + 4 (t = −2), and the limit is (−2)2 − 2 · (−2) + 4 = 12.
t+2
x2 + 6x + 5 x+5
9. After simplification, 2
= (x = −1), and the limit is (−1 + 5)/(−1 − 4) = −4/5.
x − 3x − 4 x−4
x2 − 4x + 4 x−2
10. After simplification, 2
= (x = 2), and the limit is (2 − 2)/(2 + 3) = 0.
x +x−6 x+3
x3 − 9x x(x − 3) −3(−3 − 3)
11. After simplification, 2
= (x = −3), and the limit is = −9.
x + 4x + 3 x+1 −3 + 1
3x2 − x − 2 3x + 2
12. After simplification, 2
= (x = 1), and the limit is (3 · 1 + 2)/(2 · 1 + 3) = 1.
2x + x − 3 2x + 3
t3 + 3t2 − 12t + 4 t2 + 5t − 2
13. After simplification, = (t = 2), and the limit is (22 + 5 · 2 − 2)/(22 + 2 · 2) = 3/2.
t3 − 4t t2 + 2t
t3 + t2 − 5t + 3 t+3
14. After simplification, 3
= (t = 1), and the limit is (1 + 3)/(1 + 2) = 4/3.
t − 3t + 2 t+2
15. The limit is +∞.
16. The limit is −∞.
17. The limit does not exist.
18. The limit is +∞.
19. The limit is −∞.
20. The limit does not exist.
21. The limit is +∞.
22. The limit is −∞.
23. The limit does not exist.
24. The limit is −∞.
25. The limit is +∞.
26. The limit does not exist.
27. The limit is +∞.
28. The limit is +∞.
x−9 √ √
29. After simplification, √ = x + 3 (x = 9), and the limit is 9 + 3 = 6.
x−3
Transcendentals Single Variable 12th Edition
Howard Anton
Hello all ,
We have all what you need with best price
Our email :
Our website :
testbanks-store.com
,Limits and Continuity
Exercise Set 1.1
1. (a) 2 (b) 2 (c) 2 (d) 2
2. (a) 1 (b) 1 (c) 1 (d) 1
3. (a) −1 (b) 3 (c) does not exist (d) 1
4. (a) 2 (b) 0 (c) does not exist (d) 2
5. (a) 4 (b) 4 (c) 4 (d) 2
6. (a) 1 (b) 1 (c) 1 (d) 0
7. (a) −∞ (b) −∞ (c) −∞ (d) 1
8. (a) +∞ (b) +∞ (c) +∞ (d) cannot be found from graph
9. (a) +∞ (b) +∞ (c) 2 (d) 2 (e) −∞ (f ) x = −2, x = 0, x = 2
10. (a) does not exist (b) −∞ (c) 0 (d) −1 (e) +∞ (f ) 3 (g) x = −2, x = 2
11. (i) −0.1 −0.01 −0.001 0.001 0.01 0.1
1.9866933 1.9998667 1.9999987 1.9999987 1.9998667 1.9866933
–0.1 0.1
2.01 2.01
1.99 1.99
(ii) –0.1 0.1 The limit appears to be 2.
1
,2 Chapter 1
12. (i) −0.5 −0.05 −0.005 0.005 0.05 0.5
−0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752
–0.5 0.5
–0.49 –0.49
–0.51 –0.51
(ii) –0.5 0.5 The limit appears to be −1/2.
13. (a) 2 1.5 1.1 1.01 1.001 0 0.5 0.9 0.99 0.999
0.1429 0.2105 0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.3337
0.9 1.1
0.37 0.37
0.333 0.333
0.3 0.3
0.9 1.1 The limit is 1/3.
(b) 2 1.5 1.1 1.01 1.001 1.0001
0.4286 1.0526 6.344 66.33 666.3 6666.3
1 1.1
100 100
0 0
1 1.1 The limit is +∞.
(c) 0 0.5 0.9 0.99 0.999 0.9999
−1 −1.7143 −7.0111 −67.001 −667.0 −6667.0
0.9 1
0 0
–100 –100
0.9 1 The limit is −∞.
,Exercise Set 1.1 3
14. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25
0.5359 0.5132 0.5001 0.5000 0.5000 0.4999 0.4881 0.4721
–0.1 0.1
0.51 0.51
0.5 0.5
0.49 0.49
–0.1 0.1 The limit is 1/2.
(b) 0.25 0.1 0.001 0.0001
8.4721 20.488 2000.5 20001
0 0.01
5000 5000
0 0
0 0.01 The limit is +∞.
(c) −0.25 −0.1 −0.001 −0.0001
−7.4641 −19.487 −1999.5 −20000
–0.01 0
0 0
–5000 –5000
–0.01 0 The limit is −∞.
15. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25
2.7266 2.9552 3.0000 3.0000 3.0000 3.0000 2.9552 2.7266
–0.05 0.05
3.01 3.01
2.99 2.99
–0.05 0.05 The limit is 3.
,4 Chapter 1
(b) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.001
1 1.7552 6.2161 54.87 541.1 −0.1415 −4.536 −53.19 −539.5
–1.01 –1 –0.99
2000 2000
–2000 –2000
–1.01 –1 –0.99 The limit does not exist.
16. (a) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.001
1.5574 1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.0000
–1.2 –1 –0.8
1.01 1.01
1 1
0.99 0.99
–1.2 –1 –0.8 The limit is 1.
(b) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25
1.9794 2.4132 2.5000 2.5000 2.5000 2.5000 2.4132 1.9794
–0.1 0.1
2.6 2.6
2.5 2.5
2.4 2.4
–0.1 0.1 The limit is 5/2.
17. False; define f (x) = x for x = a and f (a) = a + 1. Then limx→a f (x) = a = f (a) = a + 1.
18. True; by 1.1.3.
19. False; define f (x) = 0 for x < 0 and f (x) = x + 1 for x ≥ 0. Then the left and right limits exist but are unequal.
20. False; define f (x) = 1/x for x > 0 and f (0) = 2.
21. Answers may vary. One possible graph:
y
y = f(x)
x
,Exercise Set 1.1 5
22. Answers may vary. One possible graph:
y y = f(x)
2
1
x
–2 –1 1 2 3
23. Answers may vary. One possible graph:
y
y = f(x)
x
24. Answers may vary. One possible graph:
y
y = f(x)
x
25. Answers may vary. One possible graph:
y y = f(x)
2
x
–3 –1 1 3
–2
,6 Chapter 1
26. Answers may vary. One possible graph:
y
y = f(x)
x
x2 − 1
27. msec = = x − 1 which gets close to −2 as x gets close to −1, thus y − 1 = −2(x + 1) or y = 1 − 2(x + 1).
x+1
x2
28. msec = = x which gets close to 0 as x gets close to 0, thus y = 0.
x
x4 − 1
29. msec = = x3 + x2 + x + 1 which gets close to 4 as x gets close to 1, thus y − 1 = 4(x − 1) or y = 1 + 4(x − 1).
x−1
x4 − 1
30. msec = = x3 − x2 + x − 1 which gets close to −4 as x gets close to −1, thus y − 1 = −4(x + 1) or
x+1
y = 1 − 4(x + 1).
31. (a) The length of the rod while at rest.
(b) The limit is zero. The length of the rod approaches zero as its speed approaches c.
32. (a) The mass of the object while at rest.
(b) The limiting mass as the velocity approaches the speed of light; the mass is unbounded.
33. (a) As x → 4− , g(x) → 0+ , so lim− f (g(x)) = lim+ f (u) = 2
x→4 u→0
+ −
(b) As x → 4 , g(x) → 0 , so lim+ f (g(x)) = lim− f (u) = +∞
x→4 u→0
−
(c) As x → 0, g(x) → 2 , so lim f (g(x)) = lim f (u) = 2
x→0 u→2−
34. (a) As x → 4− , g(x) → 0+ , so lim f (g(x)) = lim f (u) = −1
x→4− u→0+
(b) As x → 4+ , g(x) → 0− , so lim f (g(x)) = lim f (u) = 0
x→4+ u→0−
−
(c) As x → 0, g(x) → 2 , so lim f (g(x)) = lim− f (u) = +∞
x→0 u→2
35. (a)
–0.001 0.001
3.5 3.5
3 3
2.5 2.5
–0.001 0.001
The limit appears to be 3.
,Exercise Set 1.2 7
(b)
–1.× 10–6 1.× 10–6
3.5 3.5
3 3
2.5 2.5
–1.× 10–6 1.× 10–6
The limit appears to not exist.
(c) Answers may vary with calculating utility. Using Mathematica:
x 0.001 (0.001)2 (0.001)3 (0.001)4 (0.001)5
f (x) 3.00416 3.00446 1.0 1.0 1.0
The limit appears to be 1.
(d) The limited precision of graphing and calculating utilities may compromise accurate calculation.
Exercise Set 1.2
1. (a) By Theorem 1.2.2, this limit is 2 + 2 · (−4) = −6.
(b) By Theorem 1.2.2, this limit is 0 − 3 · (−4) + 1 = 13.
(c) By Theorem 1.2.2, this limit is 2 · (−4) = −8.
(d) By Theorem 1.2.2, this limit is (−4)2 = 16.
√
3
(e) By Theorem 1.2.2, this limit is 6 + 2 = 2.
2 1
(f ) By Theorem 1.2.2, this limit is =− .
(−4) 2
2. (a) By Theorem 1.2.2, this limit is 0 + 0 = 0.
(b) The limit doesn’t exist because lim f doesn’t exist and lim g does.
(c) By Theorem 1.2.2, this limit is −2 + 2 = 0.
(d) By Theorem 1.2.2, this limit is 1 + 2 = 3.
(e) By Theorem 1.2.2, this limit is 0/(1 + 0) = 0.
(f ) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.
(g) The limit doesn’t exist because f (x) is not defined for 0 < x < 2.
√
(h) By Theorem 1.2.2, this limit is 1 = 1.
3. By Theorem 1.2.3, this limit is 2 · 1 · 3 = 6.
4. By Theorem 1.2.3, this limit is 23 + 2 · 2 − 5 = 7.
5. By Theorem 1.2.4, this limit is (32 − 2 · 3)/(3 + 1) = 3/4.
, 8 Chapter 1
6. By Theorem 1.2.4, this limit is (6 · 0 − 9)/(03 − 12 · 0 + 3) = −3.
x4 − 1
7. After simplification, = x2 + 1 (x = 1), and the limit is 12 + 1 = 2.
x2 − 1
t3 + 8
8. After simplification, = t2 − 2t + 4 (t = −2), and the limit is (−2)2 − 2 · (−2) + 4 = 12.
t+2
x2 + 6x + 5 x+5
9. After simplification, 2
= (x = −1), and the limit is (−1 + 5)/(−1 − 4) = −4/5.
x − 3x − 4 x−4
x2 − 4x + 4 x−2
10. After simplification, 2
= (x = 2), and the limit is (2 − 2)/(2 + 3) = 0.
x +x−6 x+3
x3 − 9x x(x − 3) −3(−3 − 3)
11. After simplification, 2
= (x = −3), and the limit is = −9.
x + 4x + 3 x+1 −3 + 1
3x2 − x − 2 3x + 2
12. After simplification, 2
= (x = 1), and the limit is (3 · 1 + 2)/(2 · 1 + 3) = 1.
2x + x − 3 2x + 3
t3 + 3t2 − 12t + 4 t2 + 5t − 2
13. After simplification, = (t = 2), and the limit is (22 + 5 · 2 − 2)/(22 + 2 · 2) = 3/2.
t3 − 4t t2 + 2t
t3 + t2 − 5t + 3 t+3
14. After simplification, 3
= (t = 1), and the limit is (1 + 3)/(1 + 2) = 4/3.
t − 3t + 2 t+2
15. The limit is +∞.
16. The limit is −∞.
17. The limit does not exist.
18. The limit is +∞.
19. The limit is −∞.
20. The limit does not exist.
21. The limit is +∞.
22. The limit is −∞.
23. The limit does not exist.
24. The limit is −∞.
25. The limit is +∞.
26. The limit does not exist.
27. The limit is +∞.
28. The limit is +∞.
x−9 √ √
29. After simplification, √ = x + 3 (x = 9), and the limit is 9 + 3 = 6.
x−3
