Section 2.3 Calculating Limits Using the Limit Laws
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 83
20.We use the difference of squares in the numerator and the difference of cubes in the denominator.
lim→1
4− 1
3− 1= lim
→1
(2− 1)(2+ 1) ( − 1)(2+ + 1) = lim
→1
( − 1)( + 1)(2+ 1) ( − 1)(2+ + 1) = lim
→1
( + 1)(2+ 1)
2+ + 1 =2(2) 3 =4
3
21. lim
→0
√9 + − 3
= lim
→0
√9 + − 3
·
√9 + + 3
√9 + + 3= lim
→0
√9 + 2
− 32
√
9 + + 3 = lim
→0
(9 + ) − 9
√
9 + + 3
= lim
→0
√
9 + + 3 = lim
→0
√ 1
9 + + 3 = 1 3 + 3= 1
6
22. lim
→2
√4 + 1 − 3
− 2 = lim
→2
√4 + 1 − 3
− 2 ·
√4 + 1 + 3
√4 + 1 + 3= lim
→2
√4 + 12
− 32 ( − 2)√
4 + 1 + 3
= lim
→2
4 + 1 − 9 ( − 2)√
4 + 1 + 3 = lim
→2
4( − 2) ( − 2)√
4 + 1 + 3
= lim
→2
√ 4
4 + 1 + 3= 4
√9 + 3=2 3
23. lim
→3
1
−1 3
− 3 = lim
→3
1
−1 3
− 3 ·3
3= lim
→3
3 −
3( − 3) = lim
→3
−1 3 = −1
9
24. lim
→0
(3 + )−1− 3−1
= lim
→0
1 3 + −1
3
= lim
→0
3 − (3 + )
(3 + )3 = lim
→0
−
(3 + )3
= lim
→0
− 1
3(3 + )
= − 1
lim→0[3(3 + )] = − 1
3(3 + 0) = −1 9
25. lim
→0
√1 + −√ 1 −
= lim
→0
√1 + −√ 1 −
·
√1 + +√ 1 −
√1 + +√
1 − = lim
→0
√1 + 2
−√
1 − 2
√
1 + +√ 1 −
= lim
→0
(1 + ) − (1 − )
√
1 + +√
1 − = lim
→0
2
√
1 + +√
1 − = lim
→0
√ 2
1 + +√ 1 −
= 2
√1 +√ 1= 2
2= 1
26. lim
→−1
2+ 2 + 1
4− 1 = lim
→−1
( + 1)2
(2+ 1)(2− 1) = lim
→−1
( + 1)2 (2+ 1)( + 1)( − 1)
= lim
→−1
+ 1
(2+ 1)( − 1)= 0 2(−2) = 0
27. lim
→16
4 −√
16 − 2 = lim
→16
(4 −√
)(4 +√
) (16 − 2)(4 +√
)= lim
→16
16 −
(16 − )(4 +√
)
= lim
→16
1
(4 +√ )= 1 16
4 +√
16 = 1 16(8) = 1
128
28. lim
→2
2− 4 + 4
4− 32− 4= lim
→2
( − 2)2
(2− 4)(2+ 1) = lim
→2
( − 2)2 ( + 2)( − 2)(2+ 1)
= lim
→2
− 2
( + 2)(2+ 1) = 0 4 · 5 = 0
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 83
20.We use the difference of squares in the numerator and the difference of cubes in the denominator.
lim→1
4− 1
3− 1= lim
→1
(2− 1)(2+ 1) ( − 1)(2+ + 1) = lim
→1
( − 1)( + 1)(2+ 1) ( − 1)(2+ + 1) = lim
→1
( + 1)(2+ 1)
2+ + 1 =2(2) 3 =4
3
21. lim
→0
√9 + − 3
= lim
→0
√9 + − 3
·
√9 + + 3
√9 + + 3= lim
→0
√9 + 2
− 32
√
9 + + 3 = lim
→0
(9 + ) − 9
√
9 + + 3
= lim
→0
√
9 + + 3 = lim
→0
√ 1
9 + + 3 = 1 3 + 3= 1
6
22. lim
→2
√4 + 1 − 3
− 2 = lim
→2
√4 + 1 − 3
− 2 ·
√4 + 1 + 3
√4 + 1 + 3= lim
→2
√4 + 12
− 32 ( − 2)√
4 + 1 + 3
= lim
→2
4 + 1 − 9 ( − 2)√
4 + 1 + 3 = lim
→2
4( − 2) ( − 2)√
4 + 1 + 3
= lim
→2
√ 4
4 + 1 + 3= 4
√9 + 3=2 3
23. lim
→3
1
−1 3
− 3 = lim
→3
1
−1 3
− 3 ·3
3= lim
→3
3 −
3( − 3) = lim
→3
−1 3 = −1
9
24. lim
→0
(3 + )−1− 3−1
= lim
→0
1 3 + −1
3
= lim
→0
3 − (3 + )
(3 + )3 = lim
→0
−
(3 + )3
= lim
→0
− 1
3(3 + )
= − 1
lim→0[3(3 + )] = − 1
3(3 + 0) = −1 9
25. lim
→0
√1 + −√ 1 −
= lim
→0
√1 + −√ 1 −
·
√1 + +√ 1 −
√1 + +√
1 − = lim
→0
√1 + 2
−√
1 − 2
√
1 + +√ 1 −
= lim
→0
(1 + ) − (1 − )
√
1 + +√
1 − = lim
→0
2
√
1 + +√
1 − = lim
→0
√ 2
1 + +√ 1 −
= 2
√1 +√ 1= 2
2= 1
26. lim
→−1
2+ 2 + 1
4− 1 = lim
→−1
( + 1)2
(2+ 1)(2− 1) = lim
→−1
( + 1)2 (2+ 1)( + 1)( − 1)
= lim
→−1
+ 1
(2+ 1)( − 1)= 0 2(−2) = 0
27. lim
→16
4 −√ 16 − 2 = lim
→16
(4 −√ )(4 +√ ) (16 − 2)(4 +√
)= lim
→16
16 −
(16 − )(4 +√
)
= lim
→16
1
(4 +√
)= 1
16 4 +√
16 = 1 16(8) = 1
128
28. lim
→2
2− 4 + 4
4− 32− 4= lim
→2
( − 2)2
(2− 4)(2+ 1) = lim
→2
( − 2)2 ( + 2)( − 2)(2+ 1)
= lim
→2
− 2
( + 2)(2+ 1) = 0 4 · 5 = 0
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
84 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 29. lim
→0
1
√
1 + −1
= lim
→0
1 −√ 1 +
√
1 + = lim
→0
1 −√ 1 +
1 +√ 1 +
√
+ 1 1 +√
1 + = lim
→0
−
√ 1 +
1 +√ 1 +
= lim
→0
√ −1 1 +
1 +√
1 + = √ −1 1 + 0
1 +√
1 + 0 = −1 2
30. lim
→−4
√2+ 9 − 5
+ 4 = lim
→−4
√2+ 9 − 5√
2+ 9 + 5 ( + 4)√
2+ 9 + 5 = lim
→−4
(2+ 9) − 25 ( + 4)√
2+ 9 + 5
= lim
→−4
2− 16 ( + 4)√
2+ 9 + 5 = lim
→−4
( + 4)( − 4) ( + 4)√
2+ 9 + 5
= lim
→−4
− 4
√2+ 9 + 5=√−4 − 4
16 + 9 + 5= −8 5 + 5= −4
5
31. lim
→0
( + )3− 3
= lim
→0
(3+ 32 + 32+ 3) − 3
= lim
→0
32 + 32+ 3
= lim
→0
(32+ 3 + 2)
= lim
→0(32+ 3 + 2) = 32
32. lim
→0
1
( + )2 − 1
2
= lim
→0
2− ( + )2 ( + )22
= lim
→0
2− (2+ 2 + 2)
2( + )2 = lim
→0
−(2 + )
2( + )2
= lim
→0
−(2 + )
2( + )2 = −2
2· 2 = − 2
3 33. (a)
lim→0
√
1 + 3 − 1 ≈ 2 3
(b)
()
−0001 0666 166 3
−0000 1 0666 616 7
−0000 01 0666 661 7
−0000 001 0666 666 2 0000 001 0666 667 2 0000 01 0666 671 7 0000 1 0666 716 7 0001 0667 166 3
The limit appears to be2 3.
(c) lim
→0
√1 + 3 − 1·
√1 + 3 + 1
√1 + 3 + 1
= lim
→0
√
1 + 3 + 1 (1 + 3) − 1 = lim
→0
√
1 + 3 + 1 3
=1 3 lim
→0
√1 + 3 + 1
[Limit Law 3]
=1 3
lim
→0(1 + 3) + lim
→01
[1 and 11]
=1 3
lim
→01 + 3 lim
→0 + 1
[1, 3, and 7]
=1 3
√1 + 3 · 0 + 1 [7 and 8]
=1
3(1 + 1) = 2 3
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
84 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 29. lim
→0
1
√
1 + −1
= lim
→0
1 −√ 1 +
√
1 + = lim
→0
1 −√ 1 +
1 +√ 1 +
√
+ 1 1 +√
1 + = lim
→0
−
√ 1 +
1 +√ 1 +
= lim
→0
√ −1 1 +
1 +√
1 + = √ −1 1 + 0
1 +√
1 + 0 = −1 2
30. lim
→−4
√2+ 9 − 5
+ 4 = lim
→−4
√2+ 9 − 5√
2+ 9 + 5 ( + 4)√
2+ 9 + 5 = lim
→−4
(2+ 9) − 25 ( + 4)√
2+ 9 + 5
= lim
→−4
2− 16 ( + 4)√
2+ 9 + 5 = lim
→−4
( + 4)( − 4) ( + 4)√
2+ 9 + 5
= lim
→−4
− 4
√2+ 9 + 5=√−4 − 4
16 + 9 + 5= −8 5 + 5= −4
5
31. lim
→0
( + )3− 3
= lim
→0
(3+ 32 + 32+ 3) − 3
= lim
→0
32 + 32+ 3
= lim
→0
(32+ 3 + 2)
= lim
→0(32+ 3 + 2) = 32
32. lim
→0
1
( + )2 − 1
2
= lim
→0
2− ( + )2 ( + )22
= lim
→0
2− (2+ 2 + 2)
2( + )2 = lim
→0
−(2 + )
2( + )2
= lim
→0
−(2 + )
2( + )2 = −2
2· 2 = − 2
3 33. (a)
lim→0
√
1 + 3 − 1 ≈ 2 3
(b)
()
−0001 0666 166 3
−0000 1 0666 616 7
−0000 01 0666 661 7
−0000 001 0666 666 2 0000 001 0666 667 2 0000 01 0666 671 7 0000 1 0666 716 7 0001 0667 166 3
The limit appears to be2 3.
(c) lim
→0
√1 + 3 − 1·
√1 + 3 + 1
√1 + 3 + 1
= lim
→0
√
1 + 3 + 1 (1 + 3) − 1 = lim
→0
√
1 + 3 + 1 3
=1 3 lim
→0
√1 + 3 + 1
[Limit Law 3]
=1 3
lim
→0(1 + 3) + lim
→01
[1 and 11]
=1 3
lim
→01 + 3 lim
→0 + 1
[1, 3, and 7]
=1 3
√1 + 3 · 0 + 1 [7 and 8]
=1
3(1 + 1) = 2 3
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 85 34. (a)
lim→0
√3 + −√ 3
≈ 029
(b)
()
−0001 0288 699 2
−0000 1 0288 677 5
−0000 01 0288 675 4
−0000 001 0288 675 2 0000 001 0288 675 1 0000 01 0288 674 9 0000 1 0288 672 7 0001 0288 651 1
The limit appears to be approximately 02887.
(c) lim
→0
√3 + −√ 3
·
√3 + +√
√ 3
3 + +√ 3
= lim
→0
(3 + ) − 3
√
3 + +√
3 = lim
→0
√ 1
3 + +√ 3
=
lim→01
lim→0
√3 + + lim
→0
√3 [Limit Laws 5 and 1]
= 1
lim
→0(3 + ) +√
3 [7 and 11]
= 1
√3 + 0 +√
3 [1, 7, and 8]
= 1
2√ 3 35.Let () = −2, () = 2cos 20and () = 2. Then
−1 ≤ cos 20 ≤ 1 ⇒ −2≤ 2cos 20 ≤ 2 ⇒ () ≤ () ≤ ().
So since lim
→0 () = lim
→0() = 0, by the Squeeze Theorem we have
lim→0() = 0.
36.Let () = −√
3+ 2, () =√
3+ 2 sin(), and () =√
3+ 2. Then
−1 ≤ sin() ≤ 1 ⇒ −√
3+ 2≤√
3+ 2sin() ≤√
3+ 2 ⇒
() ≤ () ≤ (). So since lim
→0 () = lim
→0() = 0, by the Squeeze Theorem we have lim
→0() = 0.
37.We have lim
→4(4 − 9) = 4(4) − 9 = 7 and lim
→4
2− 4 + 7
= 42− 4(4) + 7 = 7. Since 4 − 9 ≤ () ≤ 2− 4 + 7 for ≥ 0, lim→4 () = 7by the Squeeze Theorem.
38.We have lim
→1(2) = 2(1) = 2and lim
→1(4− 2+ 2) = 14− 12+ 2 = 2. Since 2 ≤ () ≤ 4− 2+ 2for all ,
lim→1() = 2by the Squeeze Theorem.
39.−1 ≤ cos(2) ≤ 1 ⇒ −4 ≤ 4cos(2) ≤ 4. Since lim
→0
−4
= 0and lim
→04 = 0, we have
lim→0
4cos(2)
= 0by the Squeeze Theorem.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
86 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
40. −1 ≤ sin() ≤ 1 ⇒ −1≤ sin()≤ 1 ⇒ √
/ ≤√ sin()≤√ . Since lim
→0+(√/) = 0and lim
→0+(√ ) = 0, we have lim
→0+
√ sin()
= 0by the Squeeze Theorem.
41. | − 3| =
− 3 if − 3 ≥ 0
−( − 3) if − 3 0 =
− 3 if ≥ 3 3 − if 3 Thus, lim
→3+(2 + | − 3|) = lim
→3+(2 + − 3) = lim
→3+(3 − 3) = 3(3) − 3 = 6 and lim
→3−(2 + | − 3|) = lim
→3−(2 + 3 − ) = lim
→3−( + 3) = 3 + 3 = 6. Since the left and right limits are equal,
lim→3(2 + | − 3|) = 6.
42. | + 6| =
+ 6 if + 6 ≥ 0
−( + 6) if + 6 0 =
+ 6 if ≥ −6
−( + 6) if −6 We’ll look at the one-sided limits.
lim
→−6+
2 + 12
| + 6| = lim
→−6+
2( + 6)
+ 6 = 2 and lim
→−6−
2 + 12
| + 6| = lim
→−6−
2( + 6)
−( + 6) = −2 The left and right limits are different, so lim
→−6
2 + 12
| + 6| does not exist.
43.
23− 2
=
2(2 − 1)
=
2
· |2 − 1| = 2|2 − 1|
|2 − 1| =
2 − 1 if 2 − 1 ≥ 0
−(2 − 1) if 2 − 1 0 =
2 − 1 if ≥ 05
−(2 − 1) if 05 So23− 2 = 2[−(2 − 1)] for 05.
Thus, lim
→05−
2 − 1
|23− 2|= lim
→05−
2 − 1
2[−(2 − 1)] = lim
→05−
−1
2 = −1
(05)2 = −1 025= −4.
44. Since || = − for 0, we have lim
→−2
2 − ||
2 + = lim
→−2
2 − (−) 2 + = lim
→−2
2 + 2 + = lim
→−21 = 1.
45. Since || = − for 0, we have lim
→0−
1
− 1
||
= lim
→0−
1
− 1
−
= lim
→0−
2
, which does not exist since the denominator approaches 0 and the numerator does not.
46. Since || = for 0, we have lim
→0+
1
− 1
||
= lim
→0+
1
−1
= lim
→0+0 = 0.
47. (a) (b) (i) Since sgn = 1 for 0, lim
→0+sgn = lim
→0+1 = 1.
(ii) Since sgn = −1 for 0, lim
→0−sgn = lim
→0−−1 = −1.
(iii) Since lim
→0−sgn 6= lim
→0+
sgn , lim
→0sgn does not exist.
(iv) Since |sgn | = 1 for 6= 0, lim
→0|sgn | = lim
→01 = 1.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
86 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
40. −1 ≤ sin() ≤ 1 ⇒ −1≤ sin()≤ 1 ⇒ √
/ ≤√
sin()≤√ . Since lim
→0+(√
/) = 0and
lim
→0+(√ ) = 0, we have lim
→0+
√ sin()
= 0by the Squeeze Theorem.
41. | − 3| =
− 3 if − 3 ≥ 0
−( − 3) if − 3 0 =
− 3 if ≥ 3 3 − if 3 Thus, lim
→3+(2 + | − 3|) = lim
→3+(2 + − 3) = lim
→3+(3 − 3) = 3(3) − 3 = 6 and lim
→3−(2 + | − 3|) = lim
→3−(2 + 3 − ) = lim
→3−
( + 3) = 3 + 3 = 6. Since the left and right limits are equal,
lim→3(2 + | − 3|) = 6.
42. | + 6| =
+ 6 if + 6 ≥ 0
−( + 6) if + 6 0 =
+ 6 if ≥ −6
−( + 6) if −6 We’ll look at the one-sided limits.
lim
→−6+
2 + 12
| + 6| = lim
→−6+
2( + 6)
+ 6 = 2 and lim
→−6−
2 + 12
| + 6| = lim
→−6−
2( + 6)
−( + 6) = −2 The left and right limits are different, so lim
→−6
2 + 12
| + 6| does not exist.
43. 23− 2 =2(2 − 1) =2
· |2 − 1| = 2|2 − 1|
|2 − 1| =
2 − 1 if 2 − 1 ≥ 0
−(2 − 1) if 2 − 1 0 =
2 − 1 if ≥ 05
−(2 − 1) if 05 So23− 2 = 2[−(2 − 1)] for 05.
Thus, lim
→05−
2 − 1
|23− 2|= lim
→05−
2 − 1
2[−(2 − 1)] = lim
→05−
−1
2 = −1
(05)2 = −1 025= −4.
44. Since || = − for 0, we have lim
→−2
2 − ||
2 + = lim
→−2
2 − (−) 2 + = lim
→−2
2 + 2 + = lim
→−21 = 1.
45. Since || = − for 0, we have lim
→0−
1
− 1
||
= lim
→0−
1
− 1
−
= lim
→0−
2
, which does not exist since the denominator approaches 0 and the numerator does not.
46. Since || = for 0, we have lim
→0+
1
− 1
||
= lim
→0+
1
−1
= lim
→0+
0 = 0.
47. (a) (b) (i) Since sgn = 1 for 0, lim
→0+sgn = lim
→0+
1 = 1.
(ii) Since sgn = −1 for 0, lim
→0−sgn = lim
→0−−1 = −1.
(iii) Since lim
→0−sgn 6= lim
→0+sgn , lim
→0sgn does not exist.
(iv) Since |sgn | = 1 for 6= 0, lim
→0|sgn | = lim
→01 = 1.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
88 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 51. For the lim
→2()to exist, the one-sided limits at = 2 must be equal. lim
→2−() = lim
→2−
4 −12
= 4 − 1 = 3 and lim
→2+() = lim
→2+
√ + =√2 + . Now 3 =√
2 + ⇒ 9 = 2 + ⇔ = 7.
52. (a) (i) lim
→1−() = lim
→1− = 1 (ii) lim
→1+() = lim
→1+(2 − 2) = 2 − 12= 1. Since lim
→1−() = 1and lim
→1+() = 1, we have lim
→1() = 1.
Note that the fact (1) = 3 does not affect the value of the limit.
(iii) When = 1, () = 3, so (1) = 3.
(iv) lim
→2−() = lim
→2−(2 − 2) = 2 − 22= 2 − 4 = −2 (v) lim
→2+() = lim
→2+( − 3) = 2 − 3 = −1 (vi) lim
→2()does not exist since lim
→2−() 6= lim
→2+
().
(b)
() =
if 1
3 if = 1
2 − 2 if 1 ≤ 2
− 3 if 2
53. (a) (i) [[]] = −2 for −2 ≤ −1, so lim
→−2+[[]] = lim
→−2+(−2) = −2 (ii) [[]] = −3 for −3 ≤ −2, so lim
→−2−[[]] = lim
→−2−(−3) = −3.
The right and left limits are different, so lim
→−2[[]]does not exist.
(iii) [[]] = −3 for −3 ≤ −2, so lim→−24[[]] = lim
→−24(−3) = −3.
(b) (i) [[]] = − 1 for − 1 ≤ , so lim
→−[[]] = lim
→−( − 1) = − 1.
(ii) [[]] = for ≤ + 1, so lim
→+[[]] = lim
→+ = .
(c) lim
→[[]]exists ⇔ is not an integer.
54. (a) See the graph of = cos .
Since −1 ≤ cos 0 on [− −2), we have = () = [[cos ]] = −1 on [− −2).
Since 0 ≤ cos 1 on [−2 0) ∪ (0 2], we have () = 0 on [−2 0) ∪ (0 2].
Since −1 ≤ cos 0 on (2 ], we have () = −1 on (2 ].
Note that (0) = 1.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
1
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 89 (b) (i) lim
→0− () = 0and lim
→0+
() = 0, so lim
→0 () = 0.
(ii) As → (2)−, () → 0, so lim
→(2)− () = 0.
(iii) As → (2)+, () → −1, so lim
→(2)+ () = −1.
(iv) Since the answers in parts (ii) and (iii) are not equal, lim
→2 ()does not exist.
(c) lim
→ ()exists for all in the open interval (− ) except = −2 and = 2.
55.The graph of () = [[]] + [[−]] is the same as the graph of () = −1 with holes at each integer, since () = 0 for any integer . Thus, lim
→2− () = −1 and lim
→2+ () = −1, so lim
→2 () = −1. However,
(2) = [[2]] + [[−2]] = 2 + (−2) = 0, so lim
→2 () 6= (2).
56. lim
→−
0
1 −2
2
= 0√
1 − 1 = 0. As the velocity approaches the speed of light, the length approaches 0.
A left-hand limit is necessary since is not defined for .
57.Since () is a polynomial, () = 0+ 1 + 22+ · · · + . Thus, by the Limit Laws,
lim→() = lim
→
0+ 1 + 22+ · · · +
= 0+ 1lim
→ + 2lim
→2+ · · · + lim
→
= 0+ 1 + 22+ · · · + = () Thus, for any polynomial , lim
→() = ().
58.Let () = ()
()where () and () are any polynomials, and suppose that () 6= 0. Then
lim→() = lim
→
()
() =
lim→()
lim→ () [Limit Law 5] =()
() [Exercise 57] = ().
59. lim
→1[ () − 8] = lim
→1
() − 8
− 1 · ( − 1)
= lim
→1
() − 8
− 1 · lim
→1( − 1) = 10 · 0 = 0.
Thus, lim
→1 () = lim
→1{[() − 8] + 8} = lim
→1[ () − 8] + lim
→18 = 0 + 8 = 8.
Note: The value of lim
→1
() − 8
− 1 does not affect the answer since it’s multiplied by 0. What’s important is that
lim→1
() − 8
− 1 exists.
60. (a) lim
→0 () = lim
→0
()
2 · 2
= lim
→0
()
2 · lim
→02= 5 · 0 = 0 (b) lim
→0
()
= lim
→0
()
2 ·
= lim
→0
()
2 · lim
→0 = 5 · 0 = 0 61.Observe that 0 ≤ () ≤ 2for all , and lim
→00 = 0 = lim
→02. So, by the Squeeze Theorem, lim
→0 () = 0.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 89 (b) (i) lim
→0− () = 0and lim
→0+ () = 0, so lim
→0 () = 0.
(ii) As → (2)−, () → 0, so lim
→(2)−
() = 0.
(iii) As → (2)+, () → −1, so lim
→(2)+ () = −1.
(iv) Since the answers in parts (ii) and (iii) are not equal, lim
→2 ()does not exist.
(c) lim
→ ()exists for all in the open interval (− ) except = −2 and = 2.
55.The graph of () = [[]] + [[−]] is the same as the graph of () = −1 with holes at each integer, since () = 0 for any integer . Thus, lim
→2− () = −1 and lim
→2+ () = −1, so lim
→2 () = −1. However,
(2) = [[2]] + [[−2]] = 2 + (−2) = 0, so lim
→2 () 6= (2).
56. lim
→−
0
1 −2
2
= 0
√1 − 1 = 0. As the velocity approaches the speed of light, the length approaches 0.
A left-hand limit is necessary since is not defined for .
57.Since () is a polynomial, () = 0+ 1 + 22+ · · · + . Thus, by the Limit Laws,
lim→() = lim
→
0+ 1 + 22+ · · · +
= 0+ 1lim
→ + 2lim
→2+ · · · + lim
→
= 0+ 1 + 22+ · · · + = () Thus, for any polynomial , lim
→() = ().
58.Let () = ()
()where () and () are any polynomials, and suppose that () 6= 0. Then
lim→() = lim
→
()
() =
lim→()
lim→ () [Limit Law 5] =()
() [Exercise 57] = ().
59. lim
→1[ () − 8] = lim
→1
() − 8
− 1 · ( − 1)
= lim
→1
() − 8
− 1 · lim
→1( − 1) = 10 · 0 = 0.
Thus, lim
→1 () = lim
→1{[() − 8] + 8} = lim
→1[ () − 8] + lim
→18 = 0 + 8 = 8.
Note: The value of lim
→1
() − 8
− 1 does not affect the answer since it’s multiplied by 0. What’s important is that
lim→1
() − 8
− 1 exists.
60. (a) lim
→0 () = lim
→0
()
2 · 2
= lim
→0
()
2 · lim
→02= 5 · 0 = 0 (b) lim
→0
()
= lim
→0
()
2 ·
= lim
→0
()
2 · lim
→0 = 5 · 0 = 0 61.Observe that 0 ≤ () ≤ 2for all , and lim
→00 = 0 = lim
→02. So, by the Squeeze Theorem, lim
→0 () = 0.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 89 (b) (i) lim
→0− () = 0and lim
→0+ () = 0, so lim
→0 () = 0.
(ii) As → (2)−, () → 0, so lim
→(2)−
() = 0.
(iii) As → (2)+, () → −1, so lim
→(2)+ () = −1.
(iv) Since the answers in parts (ii) and (iii) are not equal, lim
→2 ()does not exist.
(c) lim
→ ()exists for all in the open interval (− ) except = −2 and = 2.
55.The graph of () = [[]] + [[−]] is the same as the graph of () = −1 with holes at each integer, since () = 0 for any integer . Thus, lim
→2− () = −1 and lim
→2+ () = −1, so lim
→2 () = −1. However,
(2) = [[2]] + [[−2]] = 2 + (−2) = 0, so lim
→2 () 6= (2).
56. lim
→−
0
1 −2
2
= 0
√1 − 1 = 0. As the velocity approaches the speed of light, the length approaches 0.
A left-hand limit is necessary since is not defined for .
57.Since () is a polynomial, () = 0+ 1 + 22+ · · · + . Thus, by the Limit Laws,
lim→() = lim
→
0+ 1 + 22+ · · · +
= 0+ 1lim
→ + 2lim
→2+ · · · + lim
→
= 0+ 1 + 22+ · · · + = () Thus, for any polynomial , lim
→() = ().
58.Let () = ()
()where () and () are any polynomials, and suppose that () 6= 0. Then
lim→() = lim
→
()
() =
lim→()
lim→ () [Limit Law 5] =()
() [Exercise 57] = ().
59. lim
→1[ () − 8] = lim
→1
() − 8
− 1 · ( − 1)
= lim
→1
() − 8
− 1 · lim
→1( − 1) = 10 · 0 = 0.
Thus, lim
→1 () = lim
→1{[() − 8] + 8} = lim
→1[ () − 8] + lim
→18 = 0 + 8 = 8.
Note: The value of lim
→1
() − 8
− 1 does not affect the answer since it’s multiplied by 0. What’s important is that
lim→1
() − 8
− 1 exists.
60. (a) lim
→0 () = lim
→0
()
2 · 2
= lim
→0
()
2 · lim
→02= 5 · 0 = 0 (b) lim
→0
()
= lim
→0
()
2 ·
= lim
→0
()
2 · lim
→0 = 5 · 0 = 0 61.Observe that 0 ≤ () ≤ 2for all , and lim
→00 = 0 = lim
→02. So, by the Squeeze Theorem, lim
→0 () = 0.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
90 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
62. Let () = [[]] and () = −[[]]. Then lim→3 ()and lim
→3()do not exist [Example 10]
but lim
→3[ () + ()] = lim
→3([[]] − [[]]) = lim
→30 = 0.
63. Let () = () and () = 1 − (), where is the Heaviside function defined in Exercise 1.3.59.
Thus, either or is 0 for any value of . Then lim
→0 ()and lim
→0()do not exist, but lim
→0[ ()()] = lim
→00 = 0.
64. lim
→2
√6 − − 2
√3 − − 1 = lim
→2
√√6 − − 2 3 − − 1·
√6 − + 2
√6 − + 2·
√3 − + 1
√3 − + 1
= lim
→2
√6 − 2
− 22
√3 − 2
− 12 ·
√3 − + 1
√6 − + 2
= lim
→2
6 − − 4 3 − − 1·
√3 − + 1
√6 − + 2
= lim
→2
(2 − )√
3 − + 1 (2 − )√
6 − + 2 = lim
→2
√3 − + 1
√6 − + 2 =1 2
65. Since the denominator approaches 0 as → −2, the limit will exist only if the numerator also approaches 0as → −2. In order for this to happen, we need lim
→−2
32+ + + 3
= 0 ⇔
3(−2)2+ (−2) + + 3 = 0 ⇔ 12 − 2 + + 3 = 0 ⇔ = 15. With = 15, the limit becomes
lim→−2
32+ 15 + 18
2+ − 2 = lim
→−2
3( + 2)( + 3) ( − 1)( + 2) = lim
→−2
3( + 3)
− 1 =3(−2 + 3)
−2 − 1 = 3
−3= −1.
66. Solution 1: First, we find the coordinates of and as functions of . Then we can find the equation of the line determined by these two points, and thus find the -intercept (the point ), and take the limit as → 0. The coordinates of are (0 ).
The point is the point of intersection of the two circles 2+ 2= 2and ( − 1)2+ 2= 1. Eliminating from these equations, we get 2− 2= 1 − ( − 1)2 ⇔ 2= 1 + 2 − 1 ⇔ = 122. Substituting back into the equation of the shrinking circle to find the -coordinate, we get1
222
+ 2= 2 ⇔ 2= 2
1 −142
⇔ = 1 −142 (the positive -value). So the coordinates of are
1 22
1 −142
. The equation of the line joining and is thus
− =
1 −142−
1
22− 0 ( − 0). We set = 0 in order to find the -intercept, and get
= −
1 22
1 −142− 1 =−122
1 −142+ 1
1 −142− 1 = 2
1 −142+ 1
Now we take the limit as → 0+: lim
→0+ = lim
→0+2
1 −142+ 1
= lim
→0+2√
1 + 1= 4.
So the limiting position of is the point (4 0).
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
