Section 2.5 Continuity
98 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
43. Given 0 we need 0 so that ln whenever 0 ; that is, = ln whenever 0 . This suggests that we take = . If 0 , then ln ln = . By the definition of a limit, lim
→0+ln = −∞.
44. (a) Let be given. Since lim
→ () = ∞, there exists 1 0such that 0 | − | 1 ⇒ () + 1 − . Since
lim→() = , there exists 2 0such that 0 | − | 2 ⇒ |() − | 1 ⇒ () − 1. Let be the smaller of 1and 2. Then 0 | − | ⇒ () + () ( + 1 − ) + ( − 1) = . Thus,
lim→[ () + ()] = ∞.
(b) Let 0 be given. Since lim
→() = 0, there exists 1 0such that 0 | − | 1 ⇒
|() − | 2 ⇒ () 2. Since lim
→ () = ∞, there exists 2 0such that 0 | − | 2 ⇒
() 2. Let = min {1 2}. Then 0 | − | ⇒ () () 2
2 = , so lim
→ () () = ∞.
(c) Let 0 be given. Since lim
→() = 0, there exists 1 0such that 0 | − | 1 ⇒
|() − | −2 ⇒ () 2. Since lim
→ () = ∞, there exists 2 0such that 0 | − | 2 ⇒
() 2. (Note that 0 and 0 ⇒ 2 0.) Let = min {1 2}. Then 0 | − | ⇒
() 2 ⇒ () () 2
·
2 = , so lim
→ () () = −∞.
2.5 Continuity
1. From Definition 1, lim
→4 () = (4).
2. The graph of has no hole, jump, or vertical asymptote.
3. (a) is discontinuous at −4 since (−4) is not defined and at −2, 2, and 4 since the limit does not exist (the left and right limits are not the same).
(b) is continuous from the left at −2 since lim
→−2− () = (−2). is continuous from the right at 2 and 4 since lim
→2+ () = (2)and lim
→4+ () = (4). It is continuous from neither side at −4 since (−4) is undefined.
4. From the graph of , we see that is continuous on the intervals [−3 −2), (−2 −1), (−1 0], (0 1), and (1 3].
5. The graph of = () must have a discontinuity at
= 2and must show that lim
→2+ () = (2).
6.The graph of = () must have discontinuities at = −1 and = 4. It must show that
lim
→−1− () = (−1) and lim
→4+ () = (4).
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100 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 13. lim
→1() = lim
→12√
32+ 1 = 2 lim
→1
√32+ 1 = 2
lim→1(32+ 1) = 2 3 lim
→12+ lim
→11
= 2
3(1)2+ 1 = 2√
4 = 4 = (1) By the definition of continuity, is continuous at = 1.
14. lim
→2 () = lim
→2
34− 5 +√3
2+ 4
= 3 lim
→24− 5 lim
→2 +3
lim→2(2+ 4)
= 3(2)4− 5(2) +√3
22+ 4 = 48 − 10 + 2 = 40 = (2) By the definition of continuity, is continuous at = 2.
15. For 4, we have
lim→ () = lim
→( +√
− 4 ) = lim
→ + lim
→
√ − 4 [Limit Law 1]
= +
lim→ − lim
→4 [8, 11, and 2]
= +√
− 4 [8 and 7]
= ()
So is continuous at = for every in (4 ∞). Also, lim
→4+ () = 4 = (4), so is continuous from the right at 4.
Thus, is continuous on [4 ∞).
16. For −2, we have
lim→() = lim
→
− 1 3 + 6=
lim→( − 1)
lim→(3 + 6) [Limit Law 5]
=
lim→ − lim
→1 3 lim
→ + lim
→6 [2 1 and 3]
= − 1
3 + 6 [8 and 7]
Thus, is continuous at = for every in (−∞ −2); that is, is continuous on (−∞ −2).
17. () = 1
+ 2is discontinuous at = −2 because (−2) is undefined.
18. () =
1
+ 2 if 6= −2
1 if = −2
Here (−2) = 1, but lim
→−2− () = −∞ and lim
→−2+ () = ∞, so lim
→−2 ()does not exist and is discontinuous at −2.
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100 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 13. lim
→1() = lim
→12√
32+ 1 = 2 lim
→1
√32+ 1 = 2
lim→1(32+ 1) = 2 3 lim
→12+ lim
→11
= 2
3(1)2+ 1 = 2√
4 = 4 = (1) By the definition of continuity, is continuous at = 1.
14. lim
→2 () = lim
→2
34− 5 +√3
2+ 4
= 3 lim
→24− 5 lim→2 +3
lim→2(2+ 4)
= 3(2)4− 5(2) +√3
22+ 4 = 48 − 10 + 2 = 40 = (2) By the definition of continuity, is continuous at = 2.
15. For 4, we have
lim→ () = lim
→( +√
− 4 ) = lim
→ + lim
→
√ − 4 [Limit Law 1]
= +
lim→ − lim
→4 [8, 11, and 2]
= +√
− 4 [8 and 7]
= ()
So is continuous at = for every in (4 ∞). Also, lim
→4+
() = 4 = (4), so is continuous from the right at 4.
Thus, is continuous on [4 ∞).
16. For −2, we have
lim→() = lim
→
− 1 3 + 6=
lim→( − 1)
lim→(3 + 6) [Limit Law 5]
=
lim→ − lim
→1 3 lim
→ + lim
→6 [2 1 and 3]
= − 1
3 + 6 [8 and 7]
Thus, is continuous at = for every in (−∞ −2); that is, is continuous on (−∞ −2).
17. () = 1
+ 2is discontinuous at = −2 because (−2) is undefined.
18. () =
1
+ 2 if 6= −2
1 if = −2
Here (−2) = 1, but lim
→−2− () = −∞ and lim
→−2+ () = ∞, so lim
→−2 ()does not exist and is discontinuous at −2.
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SECTION 2.5 CONTINUITY ¤ 101 19. () =
+ 3 if ≤ −1 2 if −1 lim
→−1− () = lim
→−1−( + 3) = −1 + 3 = 2 and lim
→−1+ () = lim
→−1+2= 2−1=12. Since the left-hand and the right-hand limits of at −1 are not equal, lim→−1 ()does not exist, and
is discontinuous at −1.
20. () =
2−
2− 1 if 6= 1
1 if = 1
lim→1 () = lim
→1
2−
2− 1 = lim
→1
( − 1)
( + 1)( − 1) = lim
→1
+ 1 =1 2, but (1) = 1, so is discontinous at 1
21. () =
cos if 0
0 if = 0
1 − 2 if 0
lim→0 () = 1, but (0) = 0 6= 1, so is discontinuous at 0.
22. () =
22− 5 − 3
− 3 if 6= 3
6 if = 3
lim→3 () = lim
→3
22− 5 − 3
− 3 = lim
→3
(2 + 1)( − 3)
− 3 = lim
→3(2 + 1) = 7, but (3) = 6, so is discontinuous at 3.
23. () =2− − 2
− 2 = ( − 2)( + 1)
− 2 = + 1for 6= 2. Since lim
→2 () = 2 + 1 = 3, define (2) = 3. Then is continuous at 2.
24. () = 2− 3 + 2
2+ − 6 =( − 1)( − 2)
( + 3)( − 2) = − 1
+ 3for 6= 2. Because lim
→2 () = 2 − 1 2 + 3= 1
5, define (2) =1
5. Then is continuous at 2.
25. () =22− − 1
2+ 1 is a rational function, so it is continuous on its domain, (−∞ ∞), by Theorem 5(b).
26. () = 2+ 1
22− − 1 = 2+ 1
(2 + 1)( − 1)is a rational function, so it is continuous on its domain,
−∞ −12
∪
−12 1
∪ (1 ∞), by Theorem 5(b).
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SECTION 2.5 CONTINUITY ¤ 103 35. Because is continuous on R and√
20 − 2is continuous on its domain, −√
20 ≤ ≤√
20, the product
() = √
20 − 2is continuous on −√
20 ≤ ≤√20. The number 2 is in that domain, so is continuous at 2, and
lim→2 () = (2) = 2√ 16 = 8.
36. Because is continuous on R, sin is continuous on R, and + sin is continuous on R, the composite function
() = sin( + sin )is continuous on R, so lim
→ () = () = sin( + sin ) = sin = 0.
37. The function () = ln
5 − 2 1 +
is continuous throughout its domain because it is the composite of a logarithm function
and a rational function. For the domain of , we must have5 − 2
1 + 0, so the numerator and denominator must have the same sign, that is, the domain is (−∞ −√
5 ] ∪ (−1√
5 ]. The number 1 is in that domain, so is continuous at 1, and
lim→1 () = (1) = ln5 − 1 1 + 1= ln 2.
38. The function () = 3√
2−2−4is continuous throughout its domain because it is the composite of an exponential function, a root function, and a polynomial. Its domain is
| 2− 2 − 4 ≥ 0
=
| 2− 2 + 1 ≥ 5
=
| ( − 1)2≥ 5
=
| − 1| ≥√ 5
= (−∞ 1 −√
5 ] ∪ [1 +√ 5 ∞) The number 4 is in that domain, so is continuous at 4, and lim
→4 () = (4) = 3√16−8−4= 32= 9.
39. () =
1 − 2 if ≤ 1 ln if 1
By Theorem 5, since () equals the polynomial 1 − 2on (−∞ 1], is continuous on (−∞ 1].
By Theorem 7, since () equals the logarithm function ln on (1 ∞), is continuous on (1 ∞).
At = 1, lim
→1− () = lim
→1−(1 − 2) = 1 − 12= 0 and lim
→1+ () = lim
→1+
ln = ln 1 = 0. Thus, lim
→1 ()exists and equals 0. Also, (1) = 1 − 12= 0. Thus, is continuous at = 1. We conclude that is continuous on (−∞ ∞).
40. () =
sin if 4 cos if ≥ 4
By Theorem 7, the trigonometric functions are continuous. Since () = sin on (−∞ 4) and () = cos on (4 ∞), is continuous on (−∞ 4) ∪ (4 ∞) lim
→(4)−
() = lim
→(4)−
sin = sin4 = 1√
2since the sine function is continuous at 4 Similarly, lim
→(4)+ () = lim
→(4)+cos = 1√
2by continuity of the cosine function at 4. Thus, lim
→(4) ()exists and equals 1√
2, which agrees with the value (4). Therefore, is continuous at 4, so is continuous on (−∞ ∞).
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104 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
41. () =
2 if −1
if − 1 ≤ 1 1 if ≥ 1
is continuous on (−∞ −1), (−1 1), and (1 ∞), where it is a polynomial, a polynomial, and a rational function, respectively.
Now lim
→−1− () = lim
→−1−
2= 1and lim
→−1+ () = lim
→−1+ = −1,
so is discontinuous at −1. Since (−1) = −1, is continuous from the right at −1. Also, lim
→1− () = lim
→1− = 1 and lim
→1+ () = lim
→1+
1
= 1 = (1), so is continuous at 1.
42. () =
2 if ≤ 1 3 − if 1 ≤ 4
√ if 4
is continuous on (−∞ 1), (1 4), and (4 ∞), where it is an exponential, a polynomial, and a root function, respectively.
Now lim
→1− () = lim
→1−2= 2 and lim
→1+ () = lim
→1+(3 − ) = 2. Since (1) = 2 we have continuity at 1. Also, lim
→4− () = lim
→4−(3 − ) = −1 = (4) and lim
→4+ () = lim
→4+
√ = 2, so is discontinuous at 4, but it is continuous
from the left at 4.
43. () =
1 + 2 if ≤ 0 2 − if 0 ≤ 2 ( − 2)2 if 2
is continuous on (−∞ 0), (0 2), and (2 ∞) since it is a polynomial on each of these intervals. Now lim
→0− () = lim
→0−(1 + 2) = 1and lim
→0+ () = lim
→0+(2 − ) = 2 so is discontinuous at 0. Since (0) = 1, is continuous from the left at 0 Also, lim
→2− () = lim
→2−(2 − ) = 0
lim
→2+ () = lim
→2+( − 2)2= 0, and (2) = 0, so is continuous at 2 The only number at which is discontinuous is 0.
44. By Theorem 5, each piece of is continuous on its domain. We need to check for continuity at = .
lim
→− () = lim
→−
3 =
2 and lim
→+ () = lim
→+
2 =
2 , so lim
→ () =
2 . Since () =
2 ,
is continuous at . Therefore, is a continuous function of .
45. () =
2+ 2 if 2
3− if ≥ 2
is continuous on (−∞ 2) and (2 ∞). Now lim
→2− () = lim
→2−
2+ 2
= 4 + 4and
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SECTION 2.5 CONTINUITY ¤ 105 lim
→2+ () = lim
→2+
3−
= 8 − 2. So is continuous ⇔ 4 + 4 = 8 − 2 ⇔ 6 = 4 ⇔ =23. Thus, for to be continuous on (−∞ ∞), =23.
46. () =
2− 4
− 2 if 2
2− + 3 if 2 ≤ 3 2 − + if ≥ 3 At = 2: lim
→2− () = lim
→2−
2− 4
− 2 = lim
→2−
( + 2)( − 2)
− 2 = lim
→2−( + 2) = 2 + 2 = 4 lim
→2+ () = lim
→2+(2− + 3) = 4 − 2 + 3 We must have 4 − 2 + 3 = 4, or 4 − 2 = 1 (1).
At = 3: lim
→3− () = lim
→3−(2− + 3) = 9 − 3 + 3 lim
→3+ () = lim
→3+(2 − + ) = 6 − +
We must have 9 − 3 + 3 = 6 − + , or 10 − 4 = 3 (2).
Now solve the system of equations by adding −2 times equation (1) to equation (2).
−8 + 4 = −2 10 − 4 = 3
2 = 1
So =12. Substituting12 for in (1) gives us −2 = −1, so = 12 as well. Thus, for to be continuous on (−∞ ∞),
= = 12.
47. If and are continuous and (2) = 6, then lim
→2[3 () + () ()] = 36 ⇒ 3 lim
→2 () + lim
→2 () · lim
→2() = 36 ⇒ 3(2) + (2) · 6 = 36 ⇒ 9(2) = 36 ⇒ (2) = 4.
48. (a) () = 1
and () = 1
2, so ( ◦ )() = (()) = (12) = 1 (12) = 2.
(b) The domain of ◦ is the set of numbers in the domain of (all nonzero reals) such that () is in the domain of (also all nonzero reals). Thus, the domain is
6= 0 and 1
2 6= 0
= { | 6= 0} or (−∞ 0) ∪ (0 ∞). Since ◦ is the composite of two rational functions, it is continuous throughout its domain; that is, everywhere except = 0.
49. (a) () = 4− 1
− 1 = (2+ 1)(2− 1)
− 1 =(2+ 1)( + 1)( − 1)
− 1 = (2+ 1)( + 1) [or 3+ 2+ + 1]
for 6= 1. The discontinuity is removable and () = 3+ 2+ + 1agrees with for 6= 1 and is continuous on R.
(b) () =3− 2− 2
− 2 = (2− − 2)
− 2 = ( − 2)( + 1)
− 2 = ( + 1) [or 2+ ] for 6= 2. The discontinuity is removable and () = 2+ agrees with for 6= 2 and is continuous on R.
(c) lim
→− () = lim
→−[[sin ]] = lim
→−0 = 0and lim
→+ () = lim
→+[[sin ]] = lim
→+(−1) = −1, so lim
→ ()does not exist. The discontinuity at = is a jump discontinuity.
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108 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
63. (⇒) If is continuous at , then by Theorem 8 with () = + , we have
lim→0 ( + ) =
lim→0( + )
= ().
(⇐) Let 0. Since lim
→0 ( + ) = (), there exists 0 such that 0 || ⇒
|( + ) − ()| . So if 0 | − | , then |() − ()| = |( + ( − )) − ()| .
Thus, lim
→ () = ()and so is continuous at .
64. lim
→0sin( + ) = lim
→0(sin cos + cos sin ) = lim
→0(sin cos ) + lim
→0(cos sin )
=
lim→0sin
lim→0cos +
lim→0cos
lim→0sin
= (sin )(1) + (cos )(0) = sin 65. As in the previous exercise, we must show that lim
→0cos( + ) = cos to prove that the cosine function is continuous.
lim→0cos( + ) = lim
→0(cos cos − sin sin ) = lim
→0(cos cos ) − lim
→0(sin sin )
=
lim→0cos
lim→0cos
−
lim→0sin
lim→0sin
= (cos )(1) − (sin )(0) = cos
66. (a) Since is continuous at , lim
→ () = (). Thus, using the Constant Multiple Law of Limits, we have
lim→( )() = lim
→ () = lim
→ () = () = ( )(). Therefore, is continuous at .
(b) Since and are continuous at , lim
→ () = ()and lim
→() = (). Since () 6= 0, we can use the Quotient Law of Limits: lim
→
() = lim
→
()
() =
lim→ ()
lim→() = ()
() =
(). Thus,
is continuous at .
67. () =
0 if is rational
1 if is irrational is continuous nowhere. For, given any number and any 0, the interval ( − + ) contains both infinitely many rational and infinitely many irrational numbers. Since () = 0 or 1, there are infinitely many numbers with 0 | − | and |() − ()| = 1. Thus, lim
→ () 6= (). [In fact, lim
→ ()does not even exist.]
68. () =
0 if is rational
if is irrational is continuous at 0. To see why, note that − || ≤ () ≤ ||, so by the Squeeze Theorem
lim→0() = 0 = (0). But is continuous nowhere else. For if 6= 0 and 0, the interval ( − + ) contains both infinitely many rational and infinitely many irrational numbers. Since () = 0 or , there are infinitely many numbers with 0 | − | and |() − ()| || 2. Thus, lim
→() 6= ().
69. If there is such a number, it satisfies the equation 3+ 1 = ⇔ 3− + 1 = 0. Let the left-hand side of this equation be called (). Now (−2) = −5 0, and (−1) = 1 0. Note also that () is a polynomial, and thus continuous. So by the Intermediate Value Theorem, there is a number between −2 and −1 such that () = 0, so that = 3+ 1.
70.
3+ 22− 1+
3+ − 2 = 0 ⇒ (3+ − 2) + (3+ 22− 1) = 0. Let () denote the left side of the last equation. Since is continuous on [−1 1], (−1) = −4 0, and (1) = 2 0, there exists a in (−1 1) such that
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