Section 2.3 Calculating Limits Using the Limit Laws
2. The graphs of f and g are given. Use them to evaluate each limit, if it exists. If the limit does not exist, explain why.
(a) lim
x→2[f (x) + g(x)] (b) lim
x→0[f (x) − g(x)] (c) lim
x→−1[f (x)g(x)] (d) lim
x→3 f (x)
g(x) (e) lim
x→2[x2f (x)] (f) f (−1) + lim
x→−1g(x) 102 Chapter 2 Limits and Derivatives
4. lim
xl21 sx423xdsx215x 1 3d 5. lim
tl22
t422
2t223t 1 2 6. lim
ul22su413u 1 6 7. lim
xl8
s
1 1s3xd
s2 2 6x21x3d 8. limtl2S
t32t223t 1 52D
29. lim
xl2
Î
2x3x 2 221110. (a) What is wrong with the following equation?
x21x 26
x 22 − x 1 3 (b) In view of part (a), explain why the equation
limxl2
x21x 26 x 22 − lim
xl2 sx 1 3d is correct.
11–32 Evaluate the limit, if it exists.
11. lim
xl5
x226x 1 5
x 25 12. lim
xl23
x213x x22x 212 13. lim
xl5
x225x 1 6
x 25 14. lim
xl4
x213x x22x 212 15. lim
tl23
t229
2t217t 1 3 16. lim
xl21
2x213x 1 1 x222x 2 3 17. lim
h l0
s25 1 hd2225
h 18. lim
hl0
s2 1 hd328 h 1. Given that
xliml2 fsxd − 4 lim
xl2tsxd − 22 limxl2 hsxd − 0 find the limits that exist. If the limit does not exist, explain why.
(a) lim
xl2 f f sxd 1 5tsxdg (b) lim
xl2 ftsxdg3 (c) lim
xl2sf sxd (d) lim
xl2
3fsxd tsxd (e) lim
xl2 tsxd
hsxd (f) lim
xl2 tsxdhsxd fsxd 2. The graphs of f and t are given. Use them to evaluate each
limit, if it exists. If the limit does not exist, explain why.
(a) lim
xl2 f f sxd 1tsxdg (b) lim
xl0 f f sxd 2tsxdg (c) lim
xl21 f f sxdtsxdg (d) lim
xl3
fsxd tsxd (e) lim
xl2 fx2fsxdg (f) fs21d 1 lim
xl21tsxd y=©
0 1 1 y=ƒ
0 1 1
y y
x x
3–9 Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).
3. lim
xl3 s5x323x21x 26d
any number lies between 21 and 1, we can write.
4
21 < sin 1
x
< 1
Any inequality remains true when multiplied by a positive number. We know that
x2> 0 for all x and so, multiplying each side of the inequalities in (4) by x
2, we get
2x
2<
x2sin 1
x<
x2as illustrated by Figure 8. We know that
x l0
lim x
2− 0 and lim
x l0
s2x
2d − 0
Taking f sxd − 2x
2, tsxd − x
2sins1yxd, and hsxd − x
2in the Squeeze Theorem, we obtain
x
lim
l0x
2sin 1
x
− 0
■y=≈
y=_≈
0 x
y
FIGURE 8 y − x2 sins1yxd
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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Solution:
80 ¤ CHAPTER 2 LIMITS AND DERIVATIVES (c) lim
→2
() =
lim→2 () [Limit Law 11]
=√ 4 = 2
(d) lim
→2
3 ()
() =
lim→2[3 ()]
lim→2() [Limit Law 5]
= 3 lim
→2 ()
lim→2() [Limit Law 3]
=3(4)
−2 = −6 (e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim
→2
()
(), does not exist because the denominator approaches 0 while the numerator approaches a nonzero number.
(f) lim
→2
() ()
() =
lim→2[() ()]
lim→2 () [Limit Law 5]
=
lim→2() · lim
→2()
lim→2 () [Limit Law 4]
= −2 · 0
4 = 0
2. (a) lim
→2[ () + ()] = lim
→2 () + lim
→2() [Limit Law 1]
= −1 + 2
= 1 (b) lim
→0 ()exists, but lim
→0()does not exist, so we cannot apply Limit Law 2 to lim
→0[ () − ()].
The limit does not exist.
(c) lim
→−1[ () ()] = lim
→−1 () · lim
→−1() [Limit Law 4]
= 1 · 2
= 2 (d) lim
→3 () = 1, but lim
→3() = 0, so we cannot apply Limit Law 5 to lim
→3
()
(). The limit does not exist.
Note: lim
→3−
()
() = ∞ since () → 0+as → 3−and lim
→3+
()
() = −∞ since () → 0−as → 3+. Therefore, the limit does not exist, even as an infinite limit.
(e) lim
→2
2 ()
= lim
→22· lim
→2 () [Limit Law 4]
= 22· (−1)
= −4
(f) (−1) + lim
→−1()is undefined since (−1) is not defined.
3. lim
→−2(34+ 22− + 1) = lim
→−234+ lim
→−222− lim
→−2 + lim
→−21 [Limit Laws 1 and 2]
= 3 lim
→−24+ 2 lim
→−22− lim
→−2 + lim
→−21 [3]
= 3(−2)4+ 2(−2)2− (−2) + (1) [9, 8, and 7]
= 48 + 8 + 2 + 1 = 59
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26. Evaluate the limit, if it exists. lim
h→0
(−2+h)−1+2−1 h
Solution:
90 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 17. lim
→−2
2− − 6
32+ 5 − 2 = lim
→−2
( − 3)( + 2) (3 − 1)( + 2) = lim
→−2
− 3
3 − 1= −2 − 3 3(−2) − 1 = −5
−7=5 7
18. lim
→−5
22+ 9 − 5
2− 25 = lim
→−5
(2 − 1)( + 5) ( − 5)( + 5) = lim
→−5
2 − 1
− 5 =2(−5) − 1
−5 − 5 = −11
−10=11 10 19.Factoring 3− 27 as the difference of two cubes, we have
lim→3
3− 27
2− 9 = lim
→3
( − 3)(2+ 3 + 9) ( − 3)( + 3) = lim
→3
2+ 3 + 9
+ 3 = 32+ 3(3) + 9 3 + 3 =27
6 =9 2. 20.Factoring 3+ 1as the sum of two cubes, we have
lim→−1
+ 1
3+ 1= lim
→−1
+ 1
( + 1)(2− + 1)= lim
→−1
1
2− + 1= 1
(−1)2− (−1) + 1 =1 3.
21. lim
→0
( − 3)2− 9
= lim
→0
2− 6 + 9 − 9
= lim
→0
2− 6
= lim
→0
( − 6)
= lim
→0( − 6) = 0 − 6 = −6
22. lim
→9
9 − 3 −√
= lim
→9
9 − 3 −√
·3 +√
3 +√
= lim
→9
(9 − )(3 +√
)
9 − = lim
→9(3 +√
) = 3 +√ 9 = 6
23. 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
24. lim
→2
2 −
√ + 2 − 2= lim
→2
2 −
√ + 2 − 2·
√ + 2 + 2
√ + 2 + 2 = lim
→2
(2 − )√
+ 2 + 2
√ + 22
− 4 = lim
→2
−( − 2)√
+ 2 + 2
− 2
= lim
→2
−(√
+ 2 + 2)
= −√
4 + 2
= −4
25. 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
26. lim
→0
(−2 + )−1+ 2−1
= lim
→0
1
− 2+1 2
= lim
→0
2 + ( − 2) 2( − 2)
= lim
→0
2 + ( − 2) 2( − 2)
= lim
→0
2( − 2) = lim
→0
1
2( − 2) = 1
2(0 − 2) = −1 4
27. 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
° 2021 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
34. Evaluate the limit, if it exists. lim
h→0
1 (x+h)2−1
x2
h
Solution:
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
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1
54. Let
g(x) =
x if x < 1
3 if x = 1
2 − x2 if 1 < x ≤ 2 x − 3 if x > 2 (a) Evaluate each of the following, if it exists. (i) lim
x→1−
g(x) (ii) lim
x→1g(x) (iii) g(1) (iv) lim
x→2−
g(x) (v) lim
x→2+
g(x) (vi)
x→2limg(x)
(b) Sketch the graph of g.
Solution:
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
68. The figure shows a fixed circle C1 with equation (x − 1)2+ y2 = 1 and a shrinking circle C2 with radius r and center the origin. P is the point (0, r), Q is the upper point of intersection of the two circles, and R is the point of intersection of the line P Q and the x-axis. What happens to R as C2 shrinks, that is, as r → 0+?
104 Chapter 2 Limits and Derivatives
The intuitive definition of a limit given in Section 2.2 is inadequate for some purposes because such phrases as “x is close to 2” and “ f sxd gets closer and closer to L” are vague.
In order to be able to prove conclusively that
x l0
lim Sx31 10,000 cos 5x D − 0.0001 or lim
x l0
sin x
x− 1 we must make the definition of a limit precise.
59. If lim
xl1
fsxd 2 8
x 21 − 10, find lim
xl1 fsxd.
60. If lim
xl0
fsxd
x2 − 5, find the following limits.
(a) lim
xl0 fsxd (b) lim
xl0
fsxd x 61. If
fsxd −
H
x02 if x is rational if x is irrational prove that limx l0 fsxd − 0.62. Show by means of an example that limxla f f sxd 1tsxdg may exist even though neither limxla fsxd nor limxlatsxd exists.
63. Show by means of an example that limxla f f sxdtsxdg may exist even though neither limxla fsxd nor limxlatsxd exists.
64. Evaluate lim
xl2 s6 2 x 22 s3 2 x 21. 65. Is there a number a such that
xliml22
3x21ax 1 a 13 x21x 22
exists? If so, find the value of a and the value of the limit.
66. The figure shows a fixed circle C1 with equation
sx 2 1d21y2− 1 and a shrinking circle C2 with radius r and center the origin. P is the point s0, rd, Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the x-axis. What happens to R as C2 shrinks, that is, as r l 01?
x y
0
P Q
C™
C¡ R
52. Let
tsxd −
x 3 2 2 x2 x 23
if x , 1 if x − 1 if 1 , x < 2 if x . 2 (a) Evaluate each of the following, if it exists.
(i) lim
xl12tsxd (ii) lim
xl1tsxd (iii) ts1d
(iv) lim
xl22tsxd (v) lim
xl21tsxd (vi) lim
xl2tsxd (b) Sketch the graph of t.
53. (a) If the symbol v b denotes the greatest integer function defined in Example 10, evaluate
(i) lim
xl221v x b (ii) lim
xl22 v x b (iii) lim
xl22.4v x b (b) If n is an integer, evaluate
(i) lim
xln2v x b (ii) lim
xln1 v x b (c) For what values of a does limxlav x b exist?
54. Let fsxd − vcos x b, 2 < x < .
(a) Sketch the graph of f.
(b) Evaluate each limit, if it exists.
(i) lim
xl0 fsxd (ii) lim
xls y2d2 fsxd
(iii) lim
xls y2d1 fsxd (iv) lim
xl y2 fsxd (c) For what values of a does limxla fsxd exist?
55. If fsxd − v x b 1 v2x b, show that limxl2 fsxd exists but is not equal to fs2d.
56. In the theory of relativity, the Lorentz contraction formula L − L0s1 2 v2yc2
expresses the length L of an object as a function of its velocity v with respect to an observer, where L0 is the length of the object at rest and c is the speed of light. Find limvlc2L and interpret the result. Why is a left-hand limit necessary?
57. If p is a polynomial, show that limxla psxd − psad.
58. If r is a rational function, use Exercise 57 to show that limxla rsxd − rsad for every number a in the domain of r.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Solution:
2
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).
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SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 91
Solution 2: We add a few lines to the diagram, as shown. Note that
∠ = 90◦(subtended by diameter ). So∠ = 90◦=∠
(subtended by diameter ). It follows that∠ = ∠. Also
∠ = 90◦− ∠ = ∠ . Since 4 is isosceles, so is 4 , implying that = . As the circle 2shrinks, the point plainly approaches the origin, so the point must approach a point twice as far from the origin as , that is, the point (4 0), as above.
2.4 The Precise Definition of a Limit
1. If |() − 1| 02, then −02 () − 1 02 ⇒ 08 () 12. From the graph, we see that the last inequality is true if 07 11, so we can choose = min {1 − 07 11 − 1} = min {03 01} = 01 (or any smaller positive number).
2. If |() − 2| 05, then −05 () − 2 05 ⇒ 15 () 25. From the graph, we see that the last inequality is true if 26 38, so we can take = min {3 − 26 38 − 3} = min {04 08} = 04 (or any smaller positive number).
Note that 6= 3.
3. The leftmost question mark is the solution of √ = 16 and the rightmost, √ = 24. So the values are 162= 256and 242= 576. On the left side, we need | − 4| |256 − 4| = 144. On the right side, we need | − 4| |576 − 4| = 176.
To satisfy both conditions, we need the more restrictive condition to hold — namely, | − 4| 144. Thus, we can choose
= 144, or any smaller positive number.
4. The leftmost question mark is the positive solution of 2= 12, that is, = √1
2, and the rightmost question mark is the positive solution of 2= 32, that is, =
3
2. On the left side, we need | − 1| √12− 1
≈ 0292 (rounding down to be safe). On the right side, we need | − 1|
3
2− 1 ≈ 0224. The more restrictive of these two conditions must apply, so we choose
= 0224(or any smaller positive number).
5. From the graph, we find that = tan = 08 when ≈ 0675, so
4 − 1≈ 0675 ⇒ 1≈ 4 − 0675 ≈ 01106. Also, = tan = 12 when ≈ 0876, so4 + 2≈ 0876 ⇒ 2= 0876 −4 ≈ 00906.
Thus, we choose = 00906 (or any smaller positive number) since this is the smaller of 1and 2.
6. From the graph, we find that = 2(2+ 4) = 03when =23, so 1 − 1= 23 ⇒ 1= 13. Also, = 2(2+ 4) = 05when = 2, so 1 + 2= 2 ⇒ 2= 1. Thus, we choose = 13(or any smaller positive number) since this is the smaller of 1and 2.
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3
