l 2 s d l 2 section 2.3 calculating limits using the limit

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) limx→2

[f(x) + g(x)] (b) limx→0

[f(x)− g(x)] (c) limx→−1

[f(x)g(x)] (d) limx→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. limxl 21

sx 4 2 3xdsx 2 1 5x 1 3d

5. limt l 22

t 4 2 2

2t 2 2 3t 1 2 6. lim

ul

22 su 4 1 3u 1 6

7. limx l 8

s1 1 s3 x ds2 2 6x 2 1 x 3d 8. limt l 2

S t 2 2 2

t 3 2 3t 1 5D2

9. limx l 2

Î 2x 2 1 1

3x 2 2

1�0�. (a) What is wrong with the following equation?

x 2 1 x 2 6

x 2 2− x 1 3

(b) In view of part (a), explain why the equation

limx l

2 x 2 1 x 2 6

x 2 2− lim

x l

2 sx 1 3d

is correct.

1�1�–3�2� Evaluate the limit, if it exists.

1�1�. limx l

5 x 2 2 6x 1 5

x 2 5 1�2�. lim

x l

23

x 2 1 3x

x 2 2 x 2 12

1�3�. limx l

5 x 2 2 5x 1 6

x 2 5 1�4. lim

x l

4

x 2 1 3x

x 2 2 x 2 12

1�5. limt l

23

t 2 2 9

2t 2 1 7t 1 3 1�6. lim

x l

21 2x 2 1 3x 1 1

x 2 2 2x 2 3

1�7. limh l

0 s25 1 hd2 2 25

h 1�8. lim

h l

0 s2 1 hd3 2 8

h

1�. Given that

limx l

2 f sxd − 4 lim

x l

2 tsxd − 22 lim

x l

2 hsxd − 0

find the limits that exist. If the limit does not exist, explain why.

(a) limx l

2 f f sxd 1 5tsxdg (b) lim

x l

2 ftsxdg3

(c) limx l 2

sf sxd (d) limx l

2 3f sxdtsxd

(e) limx l

2 tsxdhsxd

(f ) limx l

2 tsxdhsxd

f sxd

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) limx l

2 f f sxd 1 tsxdg (b) lim

x l

0 f f sxd 2 tsxdg

(c) limx l

21 f f sxdtsxdg (d) lim

x l

3

f sxdtsxd

(e) limx l

2 fx 2 f sxdg (f ) f s21d 1 lim

x l

21 tsxd

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�. limx l

3 s5x 3 2 3x 2 1 x 2 6d

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 x 2 > 0 for all x and so, multiplying each side of the inequalities in (4) by x 2, we get

2x 2 < x 2 sin 1

x< x 2

as illustrated by Figure 8. We know that

limx l 0

x 2 − 0 and limx l 0

s2x 2 d − 0

Taking f sxd − 2x 2, tsxd − x 2 sins1yxd, and hsxd − x 2 in the Squeeze Theorem, we obtain

limx l 0

x 2 sin 1

x− 0 ■

y=≈

y=_≈

0 x

y

FIGURE 8 y − x 2 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).

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:

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→−2

34 + lim→−2

22 − lim→−2

+ lim→−2

1 [Limit Laws 1 and 2]

= 3 lim→−2

4 + 2 lim→−2

2 − lim→−2

+ lim→−2

1 [3]

= 3(−2)4 + 2(−2)2 − (−2) + (1) [9, 8, and 7]

= 48 + 8 + 2 + 1 = 59

c° 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.

26. Evaluate the limit, if it exists. limh→0

(−2+h)−1+2−1

h

Solution:


17. lim→−2

2 − − 632 + 5− 2 = lim

→−2(− 3)(+ 2)(3− 1)(+ 2) = lim

→−2− 33− 1 =

−2− 33(−2)− 1 =

−5−7 =

5

7

18. lim→−5

22 + 9− 52 − 25 = lim

→−5(2− 1)(+ 5)(− 5)(+ 5) = lim

→−52− 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 − 272 − 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 + 1 as the sum of two cubes, we have

lim→−1

+ 1

3 + 1= lim

→−1+ 1

(+ 1)(2 − + 1)= lim

→−11

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

−13

= −19

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


34. Evaluate the limit, if it exists. limh→0

1(x+h)2

− 1x2

h

Solution:


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 + 01 +

√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

3lim→0

√1 + 3+ 1

[Limit Law 3]

=1

3

lim→0

(1 + 3) + lim→0

1

[1 and 11]

=1

3

lim→0

1 + 3 lim→0

+ 1

[1, 3, and 7]

=1

3

√1 + 3 · 0 + 1

[7 and 8]

=1

3(1 + 1) =

2

3


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) limx→1−

g(x) (ii) limx→1

g(x) (iii) g(1) (iv) limx→2−

g(x) (v) limx→2+

g(x) (vi)

limx→2

g(x)

(b) Sketch the graph of g.

Solution:


51. For the lim→2

() to exist, the one-sided limits at = 2 must be equal. lim→2−

() = lim→2−

4− 1

2

= 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−

() = 1 and 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.


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 PQ 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

limx l 0

Sx 3 1cos 5x

10,000D − 0.0001 or limx l 0

sin x

x− 1

we must make the definition of a limit precise.

59. If limx l 1

f sxd 2 8

x 2 1− 10, find lim

x l 1 f sxd.

60. If limx l 0

f sxdx 2 − 5, find the following limits.

(a) limx l 0

f sxd (b) limx l 0

f sxd

x

61. If

f sxd − Hx 2

0

if x is rational

if x is irrational

prove that lim x l 0 f sxd − 0.

62. Show by means of an example that limx l a f f sxd 1 tsxdg may exist even though neither lim x l a f sxd nor limx l a tsxd exists.

63. Show by means of an example that limx l a f f sxd tsxdg may exist even though neither limx l a f sxd nor limx l a tsxd exists.

64. Evaluate limx l 2

s6 2 x 2 2

s3 2 x 2 1.

65. Is there a number a such that

limx l

22 3x 2 1 ax 1 a 1 3

x 2 1 x 2 2

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 1d2 1 y 2 − 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 QC™

C¡R

52. Let

tsxd −

x

3

2 2 x 2

x 2 3

if x , 1

if x − 1

if 1 , x < 2

if x . 2

(a) Evaluate each of the following, if it exists. (i) lim

x l

12 tsxd (ii) lim

x l 1 tsxd (iii) ts1d

(iv) limx l

22 tsxd (v) lim

x l 21

tsxd (vi) limx l 2

tsxd

(b) Sketch the graph of t.

53. (a) If the symbol v b denotes the greatest integer function defined in Example 10, evaluate

(i) limx l

221 v x b (ii) lim

x l

22 v x b (iii) lim

x l

22.4 v x b

(b) If n is an integer, evaluate (i) lim

x l

n2 v x b (ii) lim

x l n1 v x b

(c) For what values of a does limx l a v x b exist?

54. Let f sxd − vcos x b , 2� < x < �.

(a) Sketch the graph of f. (b) Evaluate each limit, if it exists.

(i) limx l 0

f sxd (ii) limx l

s�y2d2 f sxd

(iii) limx l

s�y2d1 f sxd (iv) lim

x l

�y2 f sxd

(c) For what values of a does limx l a f sxd exist?

55. If f sxd − v x b 1 v2x b , show that limx l 2 f sxd exists but is not equal to f s2d.

56. In the theory of relativity, the Lorentz contraction formula

L − L0 s1 2 v 2yc 2

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 limv l

c2 L and interpret the result. Why is a left-hand limit necessary?

57. If p is a polynomial, show that lim xl a psxd − psad.

58. If r is a rational function, use Exercise 57 to show that limx l a 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


62. Let () = [[]] and () = −[[]]. Then lim→3

() and lim→3

() do not exist [Example 10]

but lim→3

[() + ()] = lim→3

([[]]− [[]]) = lim→3

0 = 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→0

0 = 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

0 as → −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 = 2 and (− 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 get

1222

+ 2 = 2 ⇔ 2 = 21− 1

42 ⇔ =

1− 1

42

(the positive -value). So the coordinates of are

122

1− 1

42

. The equation of the line joining and is thus

− =

1− 142 −

122 − 0

(− 0). We set = 0 in order to find the -intercept, and get

= −122

1− 142 − 1

=− 1

22

1− 142 + 1

1− 1

42 − 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).


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 2 shrinks, 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 = 256 and

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, =

32

. On the left side, we need |− 1| 1√

2− 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, so 4

+ 2 ≈ 0876 ⇒ 2 = 0876− 4≈ 00906.

Thus, we choose = 00906 (or any smaller positive number) since this is

the smaller of 1 and 2.

6. From the graph, we find that = 2(2 + 4) = 03 when = 23

, so

1− 1 = 23⇒ 1 = 1

3. Also, = 2(2 + 4) = 05 when = 2, so

1 + 2 = 2 ⇒ 2 = 1. Thus, we choose = 13

(or any smaller positive

number) since this is the smaller of 1 and 2.


3

l 2 s d l 2 section 2.3 calculating limits using the limit

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