, Solution manual For Calculus Early
Transcendentals, 11th Edition Howard Anton
Hello all ,
We have all what you need with best price
Our email :
Our website :
testbanks-store.com
, rch 26,
arch 26, 2009
2009 06:39
06:39 ”ISM
”ISM LT
LT chapter
chapter 1”
1” Sheet
Sheet number
number 11 Page
Page number
number 26
26 black
black
CHAPTER
CHAPTER 11
Limits
Limits and
and Continuity
Limits and Continuity Continuity
EXERCISE
EXERCISE SET
SET 1.1
1.1
Exercise
1. (a) 33 Set 1.1
1. (a) (b)
(b) 33 (c)
(c) 33 (d)
(d) 33
1. (a) 30
2. (b) 3 (c) 3 (d) 3
2. (a)
(a) 0 (b)
(b) 00 (c)
(c) 00 (d)
(d) 00
2. (a) 0 (b) 0 (c) 0 (d) 0
3. (a) −1
3. (a) −1 (b)
(b) 33 (c)
(c) does
does not
not exist
exist (d)
(d) 11
3. (a) −1 (b) 3 (c) does not exist (d) 1
4.
4. (a)
(a) 22 (b)
(b) 00 (c)
(c) does
does not
not exist
exist (d)
(d) 22
4. (a) 2 (b) 0 (c) does not exist (d) 2
5.
5. (a)
(a) 00 (b)
(b) 00 (c)
(c) 00 (d)
(d) 33
5. (a) 0 (b) 0 (c) 0 (d) 3
6.
6. (a)
(a) 11 (b) 1
(b) 1 (c) 1
(c) 1 (d)
(d) 00
6. (a) 1 (b) 1 (c) 1 (d) 0
7.
7. (a)
(a) −∞
−∞ (b) −∞
(b) −∞ (c) −∞
(c) −∞ (d)
(d) 11
7. (a) −∞ (b) −∞ (c) −∞ (d) 1
8.
8. (a) +∞
(a) +∞
+∞ (b) +∞
(b) +∞
+∞ (c) +∞
(c) +∞
+∞ (d) undef
(d) can
undef
8. (a) (b) (c) (d) not be found from graph
9.
9. (a) 11
(a) +∞ (b) −∞
−∞ 2 (c) does not
not exist
exist (f ) (d) −2
−2x = 0, x = 2
9. (a) (b) +∞ (b) (c) (d) 2 (c) (e)
does−∞ x(d)
= −2,
10. (a)
10. (a) +∞
(a) does
+∞not exist (b)
(b) 11
(b)−∞ (c) 0 (c)
(c)−1does
(d) does not
(e)exist
not +∞
exist (d)
(f
(d)) 322 (g) x = −2, x = 2
11.
11. (i)
11. (i)
(i) −0.1
−0.1 −0.01
−0.01 −0.001
−0.001 0.001
0.001 0.01
0.01 0.1
0.1
−0.1 −0.01 −0.001 0.001 0.01 0.1
1.9866933
1.9866933 1.9998667
1.9998667 1.9999987
1.9999987 1.9999987
1.9999987 1.9998667
1.9998667 1.9866933
1.9866933
1.9866933 1.9998667 1.9999987 1.9999987 1.9998667 1.9866933
(ii)
(ii)
2.
2. The
The limit
limit appears
appears to
to be
be 2.
2.
1.986
-0.1 0.1
1.986
-0.1 0.1
(ii) The limit appears to be 2.
12. (i)
12.
12. (i)
(i) −0.5 −0.05 −0.005 0.005 0.05 0.5
−0.5 −0.05 −0.005 0.005 0.05 0.5
−0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752
−0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752
(ii)
(ii) -0.4896698
-0.4896698 The
The limit
limit appears
appears to be −1/2.
to be −1/2.
-0.5
-0.5
-0.5 0.5
(ii) -0.5 0.5 The limit appears to be −1/2.
13. 1
13. (a)
(a) 22 1.5
1.5 1.1
1.1 1.01
1.01 1.001
1.001 00 0.5
0.5 0.9
0.9 0.99
0.99 0.999
0.999
0.1429
0.1429 0.2105
0.2105 0.3021
0.3021 0.3300
0.3300 0.3330
0.3330 1.0000
1.0000 0.5714
0.5714 0.3690
0.3690 0.3367
0.3367 0.3337
0.3337
26
26
,2 Chapter 1
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
1
0 2
0 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
50
1 2
0 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
0 1
-50 The limit is −∞.
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
0.6
-0.25 0.25
0 The limit is 1/2.
(b) 0.25 0.1 0.001 0.0001
8.4721 20.488 2000.5 20001
,Exercise Set 1.1 3
100
0 0.25
0 The limit is +∞.
(c) −0.25 −0.1 −0.001 −0.0001
−7.4641 −19.487 −1999.5 −20000
0
-0.25 0
-100 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
3
-0.25 0.25
2 The limit is 3.
(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
60
-1.5 0
-60 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
1.5
-1.5 0
1 The limit is 1.
,4 Chapter 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
2.5
-0.25 0.25
2 The limit is 5/2.
17. False; define f (x) = x for x 6= a and f (a) = a + 1. Then limx→a f (x) = a 6= 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.
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 = −2x − 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 = 4x − 3.
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 y = −4x−3.
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.
3.5
–1 1
33. (a) 2.5 The limit appears to be 3.
3.5
– 0.001 0.001
(b) 2.5 The limit appears to be 3.
,Exercise Set 1.2 5
3.5
– 0.000001 0.000001
(c) 2.5 The limit does not exist.
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.
p
(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 33 − 3 · 32 + 9 · 3 = 27.
5. By Theorem 1.2.4, this limit is (32 − 2 · 3)/(3 + 1) = 3/4.
6. By Theorem 1.2.4, this limit is (6 · 0 − 9)/(03 − 12 · 0 + 3) = −3.
x4 − 1
7. After simplification, = x3 + x2 + x + 1, and the limit is 13 + 12 + 1 + 1 = 4.
x−1
t3 + 8
8. After simplification, = t2 − 2t + 4, and the limit is (−2)2 − 2 · (−2) + 4 = 12.
t+2
x2 + 6x + 5 x+5
9. After simplification, = , and the limit is (−1 + 5)/(−1 − 4) = −4/5.
x − 3x − 4
2 x−4
,6 Chapter 1
x2 − 4x + 4 x−2
10. After simplification, = , and the limit is (2 − 2)/(2 + 3) = 0.
x2 + x − 6 x+3
2x2 + x − 1
11. After simplification, = 2x − 1, and the limit is 2 · (−1) − 1 = −3.
x+1
3x2 − x − 2 3x + 2
12. After simplification, = , and the limit is (3 · 1 + 2)/(2 · 1 + 3) = 1.
2x + x − 3
2 2x + 3
t3 + 3t2 − 12t + 4 t2 + 5t − 2
13. After simplification, = , 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, = , and the limit is (1 + 3)/(1 + 2) = 4/3.
t − 3t + 2
3 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, and the limit is 9 + 3 = 6.
x−3
4−y √ √
30. After simplification, √ = 2 + y, and the limit is 2 + 4 = 4.
2− y
31. (a) 2 (b) 2 (c) 2
32. (a) does not exist (b) 1 (c) 4
33. True, by Theorem 1.2.2.
x2
34. False; e.g. lim = 0.
x→0 x
,Exercise Set 1.2 7
35. False; e.g. f (x) = 2x, g(x) = x, so lim f (x) = lim g(x) = 0, but lim f (x)/g(x) = 2.
x→0 x→0 x→0
36. True, by Theorem 1.2.4.
√
x+4−2 1
37. After simplification, =√ , and the limit is 1/4.
x x+4+2
√
x2 + 4 − 2 x
38. After simplification, =√ , and the limit is 0.
x 2
x +4+2
x3 − 1
39. (a) After simplification, = x2 + x + 1, and the limit is 3.
x−1
y
4
x
(b) 1
x2 − 9
40. (a) After simplification, = x − 3, and the limit is −6, so we need that k = −6.
x+3
(b) On its domain (all real numbers), f (x) = x − 3.
41. (a) Theorem 1.2.2 doesn’t apply; moreover one cannot subtract infinities.
x−1
1 1
(b) lim+ − 2 = lim+ = −∞.
x→0 x x x→0 x2
42. (a) Theorem 1.2.2 assumes that L1 and L2 are real numbers, not infinities. It is in general not true that ”∞·0 = 0 ”.
x2
1 2 1 1 2 1
(b) − 2 = = for x 6= 0, so that lim − = .
x x + 2x x(x2 + 2x) x+2 x→0 x x2 + 2x 2
1 a x+1−a
43. For x 6= 1, − 2 = and for this to have a limit it is necessary that lim (x + 1 − a) = 0, i.e.
x−1 x −1 x2 − 1 x→1
1 2 x+1−2 x−1 1 1 1
a = 2. For this value, − = = 2 = and lim = .
x − 1 x2 − 1 x2 − 1 x −1 x+1 x→1 x + 1 2
44. (a) For small x, 1/x2 is much bigger than ±1/x.
1 1 x+1
(b) + 2 = . Since the numerator has limit 1 and x2 tends to zero from the right, the limit is +∞.
x x x2
45. The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any preassigned real
number. For example, let q(x) = x − x0 and let p(x) = a(x − x0 )n where n takes on the values 0, 1, 2.
46. If on the contrary lim g(x) did exist then by Theorem 1.2.2 so would lim [f (x) + g(x)], and that would be a
x→a x→a
contradiction.
47. Clearly, g(x) = [f (x) + g(x)] − f (x). By Theorem 1.2.2, lim [f (x) + g(x)] − lim f (x) = lim [f (x) + g(x) − f (x)] =
x→a x→a x→a
lim g(x).
x→a
, 8 Chapter 1
f (x) f (x) f (x)
48. By Theorem 1.2.2, lim f (x) = lim lim g(x) = lim · 0 = 0, since lim exists.
x→a x→a g(x) x→a x→a g(x) x→a g(x)
Exercise Set 1.3
1. (a) −∞ (b) +∞
2. (a) 2 (b) 0
3. (a) 0 (b) −1
4. (a) does not exist (b) 0
5. (a) 3 + 3 · (−5) = −12 (b) 0 − 4 · (−5) + 1 = 21 (c) 3 · (−5) = −15 (d) (−5)2 = 25
√
(e) 3
5+3=2 (f ) 3/(−5) = −3/5 (g) 0
(h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.
√
6. (a) 2 · 7 − (−6) = 20 (b) 6 · 7 + 7 · (−6) = 0 (c) +∞ (d) −∞ (e) 3 −42
(f ) −6/7 (g) 7 (h) −7/12
7. x 10 100 1000 10000 100000 1000000
f (x) 0.953463 0.995037 0.999500 0.999950 0.999995 0.9999995
The limit appears to be 1.
8. x −10 −100 −1000 −10000 −100000 −1000000
f (x) −1.05409255 −1.00503781 −1.00050037 −1.00005000 −1.0000050 −1.00000050
The limit appears to be −1.
9. The limit is −∞, by the highest degree term.
10. The limit is +∞, by the highest degree term.
11. The limit is +∞.
12. The limit is +∞.
13. The limit is 3/2, by the highest degree terms.
14. The limit is 5/2, by the highest degree terms.
15. The limit is 0, by the highest degree terms.
16. The limit is 0, by the highest degree terms.
17. The limit is 0, by the highest degree terms.
18. The limit is 5/3, by the highest degree terms.
19. The limit is −∞, by the highest degree terms.
20. The limit is +∞, by the highest degree terms.
21. The limit is −1/7, by the highest degree terms.
22. The limit is 4/7, by the highest degree terms.
p √
23. The limit is 3 −5/8 = − 3 5 /2, by the highest degree terms.
Transcendentals, 11th Edition Howard Anton
Hello all ,
We have all what you need with best price
Our email :
Our website :
testbanks-store.com
, rch 26,
arch 26, 2009
2009 06:39
06:39 ”ISM
”ISM LT
LT chapter
chapter 1”
1” Sheet
Sheet number
number 11 Page
Page number
number 26
26 black
black
CHAPTER
CHAPTER 11
Limits
Limits and
and Continuity
Limits and Continuity Continuity
EXERCISE
EXERCISE SET
SET 1.1
1.1
Exercise
1. (a) 33 Set 1.1
1. (a) (b)
(b) 33 (c)
(c) 33 (d)
(d) 33
1. (a) 30
2. (b) 3 (c) 3 (d) 3
2. (a)
(a) 0 (b)
(b) 00 (c)
(c) 00 (d)
(d) 00
2. (a) 0 (b) 0 (c) 0 (d) 0
3. (a) −1
3. (a) −1 (b)
(b) 33 (c)
(c) does
does not
not exist
exist (d)
(d) 11
3. (a) −1 (b) 3 (c) does not exist (d) 1
4.
4. (a)
(a) 22 (b)
(b) 00 (c)
(c) does
does not
not exist
exist (d)
(d) 22
4. (a) 2 (b) 0 (c) does not exist (d) 2
5.
5. (a)
(a) 00 (b)
(b) 00 (c)
(c) 00 (d)
(d) 33
5. (a) 0 (b) 0 (c) 0 (d) 3
6.
6. (a)
(a) 11 (b) 1
(b) 1 (c) 1
(c) 1 (d)
(d) 00
6. (a) 1 (b) 1 (c) 1 (d) 0
7.
7. (a)
(a) −∞
−∞ (b) −∞
(b) −∞ (c) −∞
(c) −∞ (d)
(d) 11
7. (a) −∞ (b) −∞ (c) −∞ (d) 1
8.
8. (a) +∞
(a) +∞
+∞ (b) +∞
(b) +∞
+∞ (c) +∞
(c) +∞
+∞ (d) undef
(d) can
undef
8. (a) (b) (c) (d) not be found from graph
9.
9. (a) 11
(a) +∞ (b) −∞
−∞ 2 (c) does not
not exist
exist (f ) (d) −2
−2x = 0, x = 2
9. (a) (b) +∞ (b) (c) (d) 2 (c) (e)
does−∞ x(d)
= −2,
10. (a)
10. (a) +∞
(a) does
+∞not exist (b)
(b) 11
(b)−∞ (c) 0 (c)
(c)−1does
(d) does not
(e)exist
not +∞
exist (d)
(f
(d)) 322 (g) x = −2, x = 2
11.
11. (i)
11. (i)
(i) −0.1
−0.1 −0.01
−0.01 −0.001
−0.001 0.001
0.001 0.01
0.01 0.1
0.1
−0.1 −0.01 −0.001 0.001 0.01 0.1
1.9866933
1.9866933 1.9998667
1.9998667 1.9999987
1.9999987 1.9999987
1.9999987 1.9998667
1.9998667 1.9866933
1.9866933
1.9866933 1.9998667 1.9999987 1.9999987 1.9998667 1.9866933
(ii)
(ii)
2.
2. The
The limit
limit appears
appears to
to be
be 2.
2.
1.986
-0.1 0.1
1.986
-0.1 0.1
(ii) The limit appears to be 2.
12. (i)
12.
12. (i)
(i) −0.5 −0.05 −0.005 0.005 0.05 0.5
−0.5 −0.05 −0.005 0.005 0.05 0.5
−0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752
−0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752
(ii)
(ii) -0.4896698
-0.4896698 The
The limit
limit appears
appears to be −1/2.
to be −1/2.
-0.5
-0.5
-0.5 0.5
(ii) -0.5 0.5 The limit appears to be −1/2.
13. 1
13. (a)
(a) 22 1.5
1.5 1.1
1.1 1.01
1.01 1.001
1.001 00 0.5
0.5 0.9
0.9 0.99
0.99 0.999
0.999
0.1429
0.1429 0.2105
0.2105 0.3021
0.3021 0.3300
0.3300 0.3330
0.3330 1.0000
1.0000 0.5714
0.5714 0.3690
0.3690 0.3367
0.3367 0.3337
0.3337
26
26
,2 Chapter 1
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
1
0 2
0 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
50
1 2
0 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
0 1
-50 The limit is −∞.
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
0.6
-0.25 0.25
0 The limit is 1/2.
(b) 0.25 0.1 0.001 0.0001
8.4721 20.488 2000.5 20001
,Exercise Set 1.1 3
100
0 0.25
0 The limit is +∞.
(c) −0.25 −0.1 −0.001 −0.0001
−7.4641 −19.487 −1999.5 −20000
0
-0.25 0
-100 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
3
-0.25 0.25
2 The limit is 3.
(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
60
-1.5 0
-60 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
1.5
-1.5 0
1 The limit is 1.
,4 Chapter 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
2.5
-0.25 0.25
2 The limit is 5/2.
17. False; define f (x) = x for x 6= a and f (a) = a + 1. Then limx→a f (x) = a 6= 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.
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 = −2x − 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 = 4x − 3.
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 y = −4x−3.
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.
3.5
–1 1
33. (a) 2.5 The limit appears to be 3.
3.5
– 0.001 0.001
(b) 2.5 The limit appears to be 3.
,Exercise Set 1.2 5
3.5
– 0.000001 0.000001
(c) 2.5 The limit does not exist.
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.
p
(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 33 − 3 · 32 + 9 · 3 = 27.
5. By Theorem 1.2.4, this limit is (32 − 2 · 3)/(3 + 1) = 3/4.
6. By Theorem 1.2.4, this limit is (6 · 0 − 9)/(03 − 12 · 0 + 3) = −3.
x4 − 1
7. After simplification, = x3 + x2 + x + 1, and the limit is 13 + 12 + 1 + 1 = 4.
x−1
t3 + 8
8. After simplification, = t2 − 2t + 4, and the limit is (−2)2 − 2 · (−2) + 4 = 12.
t+2
x2 + 6x + 5 x+5
9. After simplification, = , and the limit is (−1 + 5)/(−1 − 4) = −4/5.
x − 3x − 4
2 x−4
,6 Chapter 1
x2 − 4x + 4 x−2
10. After simplification, = , and the limit is (2 − 2)/(2 + 3) = 0.
x2 + x − 6 x+3
2x2 + x − 1
11. After simplification, = 2x − 1, and the limit is 2 · (−1) − 1 = −3.
x+1
3x2 − x − 2 3x + 2
12. After simplification, = , and the limit is (3 · 1 + 2)/(2 · 1 + 3) = 1.
2x + x − 3
2 2x + 3
t3 + 3t2 − 12t + 4 t2 + 5t − 2
13. After simplification, = , 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, = , and the limit is (1 + 3)/(1 + 2) = 4/3.
t − 3t + 2
3 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, and the limit is 9 + 3 = 6.
x−3
4−y √ √
30. After simplification, √ = 2 + y, and the limit is 2 + 4 = 4.
2− y
31. (a) 2 (b) 2 (c) 2
32. (a) does not exist (b) 1 (c) 4
33. True, by Theorem 1.2.2.
x2
34. False; e.g. lim = 0.
x→0 x
,Exercise Set 1.2 7
35. False; e.g. f (x) = 2x, g(x) = x, so lim f (x) = lim g(x) = 0, but lim f (x)/g(x) = 2.
x→0 x→0 x→0
36. True, by Theorem 1.2.4.
√
x+4−2 1
37. After simplification, =√ , and the limit is 1/4.
x x+4+2
√
x2 + 4 − 2 x
38. After simplification, =√ , and the limit is 0.
x 2
x +4+2
x3 − 1
39. (a) After simplification, = x2 + x + 1, and the limit is 3.
x−1
y
4
x
(b) 1
x2 − 9
40. (a) After simplification, = x − 3, and the limit is −6, so we need that k = −6.
x+3
(b) On its domain (all real numbers), f (x) = x − 3.
41. (a) Theorem 1.2.2 doesn’t apply; moreover one cannot subtract infinities.
x−1
1 1
(b) lim+ − 2 = lim+ = −∞.
x→0 x x x→0 x2
42. (a) Theorem 1.2.2 assumes that L1 and L2 are real numbers, not infinities. It is in general not true that ”∞·0 = 0 ”.
x2
1 2 1 1 2 1
(b) − 2 = = for x 6= 0, so that lim − = .
x x + 2x x(x2 + 2x) x+2 x→0 x x2 + 2x 2
1 a x+1−a
43. For x 6= 1, − 2 = and for this to have a limit it is necessary that lim (x + 1 − a) = 0, i.e.
x−1 x −1 x2 − 1 x→1
1 2 x+1−2 x−1 1 1 1
a = 2. For this value, − = = 2 = and lim = .
x − 1 x2 − 1 x2 − 1 x −1 x+1 x→1 x + 1 2
44. (a) For small x, 1/x2 is much bigger than ±1/x.
1 1 x+1
(b) + 2 = . Since the numerator has limit 1 and x2 tends to zero from the right, the limit is +∞.
x x x2
45. The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any preassigned real
number. For example, let q(x) = x − x0 and let p(x) = a(x − x0 )n where n takes on the values 0, 1, 2.
46. If on the contrary lim g(x) did exist then by Theorem 1.2.2 so would lim [f (x) + g(x)], and that would be a
x→a x→a
contradiction.
47. Clearly, g(x) = [f (x) + g(x)] − f (x). By Theorem 1.2.2, lim [f (x) + g(x)] − lim f (x) = lim [f (x) + g(x) − f (x)] =
x→a x→a x→a
lim g(x).
x→a
, 8 Chapter 1
f (x) f (x) f (x)
48. By Theorem 1.2.2, lim f (x) = lim lim g(x) = lim · 0 = 0, since lim exists.
x→a x→a g(x) x→a x→a g(x) x→a g(x)
Exercise Set 1.3
1. (a) −∞ (b) +∞
2. (a) 2 (b) 0
3. (a) 0 (b) −1
4. (a) does not exist (b) 0
5. (a) 3 + 3 · (−5) = −12 (b) 0 − 4 · (−5) + 1 = 21 (c) 3 · (−5) = −15 (d) (−5)2 = 25
√
(e) 3
5+3=2 (f ) 3/(−5) = −3/5 (g) 0
(h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.
√
6. (a) 2 · 7 − (−6) = 20 (b) 6 · 7 + 7 · (−6) = 0 (c) +∞ (d) −∞ (e) 3 −42
(f ) −6/7 (g) 7 (h) −7/12
7. x 10 100 1000 10000 100000 1000000
f (x) 0.953463 0.995037 0.999500 0.999950 0.999995 0.9999995
The limit appears to be 1.
8. x −10 −100 −1000 −10000 −100000 −1000000
f (x) −1.05409255 −1.00503781 −1.00050037 −1.00005000 −1.0000050 −1.00000050
The limit appears to be −1.
9. The limit is −∞, by the highest degree term.
10. The limit is +∞, by the highest degree term.
11. The limit is +∞.
12. The limit is +∞.
13. The limit is 3/2, by the highest degree terms.
14. The limit is 5/2, by the highest degree terms.
15. The limit is 0, by the highest degree terms.
16. The limit is 0, by the highest degree terms.
17. The limit is 0, by the highest degree terms.
18. The limit is 5/3, by the highest degree terms.
19. The limit is −∞, by the highest degree terms.
20. The limit is +∞, by the highest degree terms.
21. The limit is −1/7, by the highest degree terms.
22. The limit is 4/7, by the highest degree terms.
p √
23. The limit is 3 −5/8 = − 3 5 /2, by the highest degree terms.
