Section 2.2 The Limit of a Function
4. Use the given graph of f to state the value of each quantity, if it exists. If it does not exist, explain why.
(a) lim
x→2−f (x) (b) lim
x→2+f (x) (c) lim
x→2f (x) (d) f (2) (e) lim
x→4f (x) (f) f (4) 92 Chapter 2 Limits and Derivatives
1. Explain in your own words what is meant by the equation
xliml2 fsxd − 5
Is it possible for this statement to be true and yet fs2d − 3?
Explain.
2. Explain what it means to say that
xliml12fsxd − 3 and lim
x l11 fsxd − 7 In this situation is it possible that limxl1 fsxd exists?
Explain.
3. Explain the meaning of each of the following.
(a) lim
xl23fsxd − ` (b) lim
xl41fsxd − 2`
4. Use the given graph of f to state the value of each quantity, if it exists. If it does not exist, explain why.
(a) lim
xl22fsxd (b) lim
xl21fsxd (c) lim
xl2 fsxd
(d) fs2d (e) lim
xl4 fsxd (f) fs4d y
0 2 4 x
4 2
5. For the function f whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.
(a) lim
xl1 fsxd (b) lim
xl32fsxd (c) lim
xl31fsxd (d) lim
xl3 fsxd (e) fs3d y
0 2 4 x
4 2
6. For the function h whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.
(a) lim
xl232hsxd (b) lim
xl231hsxd (c) lim
xl23hsxd
(d) hs23d (e) lim
xl02 hsxd (f) lim
xl01 hsxd (g) lim
xl0 hsxd (h) hs0d (i) lim
xl2 hsxd
(j) hs2d (k) lim
xl51hsxd (l) lim
xl52 hsxd y
0 2 x
_2
_4 4 6
7. For the function t whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.
(a) lim
tl02tstd (b) lim
tl01tstd (c) lim
tl0tstd (d) lim
tl22tstd (e) lim
tl21tstd (f) lim
tl2tstd
(g) ts2d (h) lim
tl4tstd y
2 4 t 4
2
8. For the function A whose graph is shown, state the following.
(a) lim
x l23 Asxd (b) lim
x l22 Asxd (c) lim
x l21 Asxd (d) lim
x l21 Asxd (e) The equations of the vertical asymptotes
0 y
2 x
_3 5
9. For the function f whose graph is shown, state the following.
(a) lim
x l27 fsxd (b) lim
x l23 fsxd (c) lim
xl0 fsxd (d) lim
xl62fsxd (e) lim
xl61fsxd
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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:
70 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
2.2 The Limit of a Function
1. As approaches 2, () approaches 5. [Or, the values of () can be made as close to 5 as we like by taking sufficiently close to 2 (but 6= 2).] Yes, the graph could have a hole at (2 5) and be defined such that (2) = 3.
2. As approaches 2 from the left, () approaches 1; and as approaches 2 from the right, () approaches 5. No, the limit does not exist because the left- and right-hand limits are different.
3. (a) lim→−5 () = ∞ means that the values of () can be made arbitrarily large (as large as we please) by taking sufficiently close to −5 (but not equal to −5).
(b) lim→3+ () = −∞ means that the values of () can be made arbitrarily large negative by taking sufficiently close to 3 through values larger than 3.
4. (a) As approaches 2 from the left, the values of () approach 3, so lim
→2−
() = 3.
(b) As approaches 2 from the right, the values of () approach 1, so lim
→2+
() = 1.
(c) lim
→2 ()does not exist since the left-hand limit does not equal the right-hand limit.
(d) When = 2, = 3, so (2) = 3.
(e) As approaches 4, the values of () approach 4, so lim
→4 () = 4.
(f ) There is no value of () when = 4, so (4) does not exist.
5. (a) As approaches 1, the values of () approach 2, so lim
→1 () = 2.
(b) As approaches 3 from the left, the values of () approach 1, so lim
→3− () = 1.
(c) As approaches 3 from the right, the values of () approach 4, so lim
→3+ () = 4.
(d) lim
→3 ()does not exist since the left-hand limit does not equal the right-hand limit.
(e) When = 3, = 3, so (3) = 3.
6. (a) () approaches 4 as approaches −3 from the left, so lim
→−3−() = 4.
(b) () approaches 4 as approaches −3 from the right, so lim
→−3+() = 4.
(c) lim
→−3() = 4because the limits in part (a) and part (b) are equal.
(d) (−3) is not defined, so it doesn’t exist.
(e) () approaches 1 as approaches 0 from the left, so lim
→0−
() = 1.
(f ) () approaches −1 as approaches 0 from the right, so lim
→0+() = −1.
(g) lim
→0()does not exist because the limits in part (e) and part (f ) are not equal.
(h) (0) = 1 since the point (0 1) is on the graph of .
(i) Since lim
→2−() = 2and lim
→2+
() = 2, we have lim
→2() = 2.
(j) (2) is not defined, so it doesn’t exist.
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10. A patient receives a 150-mg injection of a drug every 4 hours. The graph shows the amount f (t) of the drug in the bloodstream after t hours. Find lim
t→12−
f (t) and lim
t→12+
f (t) and explain the significance of these one-sided limits.
Section2.2 The Limit of a Function 93 (f) The equations of the vertical asymptotes.
x y
0 6
_3 _7
10. A patient receives a 150-mg injection of a drug every 4 hours. The graph shows the amount fstd of the drug in the blood stream after t hours. Find
tllim122 fstd and lim
tl121 fstd
and explain the significance of these one-sided limits.
4 8 12 16 t
f(t)
150
0 300
11–12 Sketch the graph of the function and use it to determine the values of a for which limxla fsxd exists.
11. fsxd −
H
1 1 xx2 2 x2 if x , 21if 21 < x , 1 if x > 112. fsxd −
H
1 1 sin xcos xsin x if x , 0if 0 < x < if x .13–14 Use the graph of the function f to state the value of each limit, if it exists. If it does not exist, explain why.
(a) lim
xl02fsxd (b) lim
xl01fsxd (c) lim
xl0fsxd 13. fsxd − 1
1 1 e1yx 14. fsxd − x21x sx31x2 15–18 Sketch the graph of an example of a function f that satisfies all of the given conditions.
15. lim
xl02 fsxd − 21, lim
xl01 fsxd − 2, f s0d − 1 16. lim
xl0 fsxd − 1, lim
xl32 fsxd − 22, lim
xl31 fsxd − 2, fs0d − 21, f s3d − 1
;
17. lim
xl31 fsxd − 4, lim
xl32 fsxd − 2, lim
xl22 fsxd − 2, fs3d − 3, f s22d − 1
18. lim
xl02 fsxd − 2, lim
xl01 fsxd − 0, lim
xl42 fsxd − 3,
xliml41 fsxd − 0, fs0d − 2, fs4d − 1
19–22 Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).
19. lim
xl3
x223x x229,
x − 3.1, 3.05, 3.01, 3.001, 3.0001, 2.9, 2.95, 2.99, 2.999, 2.9999
20. lim
xl23
x223x x229,
x − 22.5, 22.9, 22.95, 22.99, 22.999, 22.9999, 23.5, 23.1, 23.05, 23.01, 23.001, 23.0001
21. lim
tl0
e5t21
t , t − 60.5, 60.1, 60.01, 60.001, 60.0001 22. lim
hl0
s2 1 hd5232
h ,
h − 60.5, 60.1, 60.01, 60.001, 60.0001
23–28 Use a table of values to estimate the value of the limit.
If you have a graphing device, use it to confirm your result graphically.
23. lim
xl4
ln x 2 ln 4
x 24 24. lim
pl21
1 1 p9 1 1 p15 25. lim
l0
sin 3
tan 2 26. lim
tl0
5t21 t 27. lim
x l01 xx 28. lim
x l01 x2 ln x
29. (a) By graphing the function fsxd − scos 2x 2 cos xdyx2 and zooming in toward the point where the graph crosses the y-axis, estimate the value of limx l0 fsxd.
(b) Check your answer in part (a) by evaluating fsxd for values of x that approach 0.
30. (a) Estimate the value of
xliml0
sin x sin x
by graphing the function fsxd − ssin xdyssin xd.
State your answer correct to two decimal places.
(b) Check your answer in part (a) by evaluating fsxd for values of x that approach 0.
;
;
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:
SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 71 (k) () approaches 3 as approaches 5 from the right, so lim
→5+() = 3.
(l) () does not approach any one number as approaches 5 from the left, so lim
→5−()does not exist.
7. (a) lim
→0−() = −1 (b) lim
→0+() = −2 (c) lim
→0()does not exist because the limits in part (a) and part (b) are not equal.
(d) lim
→2−() = 2 (e) lim
→2+() = 0 (f ) lim
→2()does not exist because the limits in part (d) and part (e) are not equal.
(g) (2) = 1 (h) lim
→4() = 3 8. (a) lim
→−3() = ∞ (b) lim
→2−() = −∞
(c) lim
→2+() = ∞ (d) lim
→−1() = −∞
(e) The equations of the vertical asymptotes are = −3, = −1 and = 2.
9. (a) lim
→−7 () = −∞ (b) lim
→−3 () = ∞ (c) lim
→0 () = ∞ (d) lim
→6− () = −∞ (e) lim
→6+ () = ∞
(f ) The equations of the vertical asymptotes are = −7, = −3, = 0, and = 6.
10. lim
→12− () = 150mg and lim
→12+
() = 300mg. These limits show that there is an abrupt change in the amount of drug in the patient’s bloodstream at = 12 h. The left-hand limit represents the amount of the drug just before the fourth injection.
The right-hand limit represents the amount of the drug just after the fourth injection.
11. From the graph of
() =
1 + if −1
2 if −1 ≤ 1 2 − if ≥ 1
,
we see that lim
→ ()exists for all except = −1. Notice that the right and left limits are different at = −1.
12.From the graph of
() =
1 + sin if 0 cos if 0 ≤ ≤ sin if
,
we see that lim
→ ()exists for all except = . Notice that the right and left limits are different at = .
° 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
16. Sketch the graph of an example of a function f that satisfies all of the given conditions.
x→0limf (x) = 4, lim
x→8−f (x) = 1, lim
x→8+f (x) = −3, f (0) = 6, f (8) = −1 Solution:
80 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
11. From the graph of we see that lim
→0− () = 1, but lim
→0+ () = −1, so
lim→ ()does not exist for = 0. However, lim
→ ()exists for all other values of . Thus, lim
→ ()exists for all in (−∞ 0) ∪ (0 ∞).
12. From the graph of we see that lim
→2−
() = 2, but lim
→2+
() = 1, so
lim→ ()does not exist for = 2. However, lim
→ ()exists for all other values of . Thus, lim
→ ()exists for all in (−∞ 2) ∪ (2 ∞).
13. (a) From the graph, lim
→0− () = −1.
() = √ 1 + −2 (b) From the graph, lim
→0+
() = 1.
(c) Since lim
→0− () 6= lim
→0+
(), lim
→0 ()does not exist.
14. (a) From the graph, lim
→0− () = −2.
() = 1− 2
1+ 1 (b) From the graph, lim
→0+
() = 1.
(c) Since lim
→0− () 6= lim
→0+ (), lim
→0 ()does not exist.
15. lim
→1−
() = 3, lim
→1+
() = 0, (1) = 2
16. lim
→0 () = 4, lim
→8− () = 1, lim
→8+ () = −3,
(0) = 6, (8) = −1
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38. Determine the infinite limit. lim
x→3− x2+4x x2−2x−3
Solution:
SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 83
28.For () = 2ln :
()
01 −0023 026 001 −0000 461 0001 −0000 007 00001 −0000 000
It appears that lim
→0+ () = 0.
The graph confirms that result.
29. lim
→5+
+ 1
− 5 = ∞ since the numerator is positive and the denominator approaches 0 from the positive side as → 5+.
30. lim
→5−
+ 1
− 5= −∞ since the numerator is positive and the denominator approaches 0 from the negative side as → 5−.
31. lim
→2
2
( − 2)2 = ∞ since the numerator is positive and the denominator approaches 0 through positive values as → 2.
32. lim
→3−
√
( − 3)5 = −∞ since the numerator is positive and the denominator approaches 0 from the negative side as → 3−. 33. lim
→1+ln(√
− 1) = −∞ since√
− 1 → 0+as → 1+.
34. lim
→0+ln(sin ) = −∞ since sin → 0+as → 0+.
35. lim
→(2)+
1
sec = −∞ since1
is positive and sec → −∞ as → (2)+. 36. lim
→− cot = −∞ since is positive and cot → −∞ as → −.
37. lim
→1
2+ 2
2− 2 + 1 = lim
→1
2+ 2
( − 1)2 = ∞ since the numerator is positive and the denominator approaches 0 through positive values as → 1.
38. lim
→3−
2+ 4
2− 2 − 3= lim
→3−
2+ 4
( − 3)( + 1)= −∞ since the numerator is positive and the denominator approaches 0 through negative values as → 3−.
39. lim
→0(ln 2− −2) = −∞ since ln 2→ −∞ and −2→ ∞ as → 0.
40. lim
→0+
1
− ln
= ∞ since 1
→ ∞ and ln → −∞ as → 0+.
41.The denominator of () = − 1
2 + 4is equal to 0 when = −2 (and the numerator is not), so = −2 is the vertical asymptote of the function.
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42. (a) Find the vertical asymptotes of the function y = 3x−2xx2+12
(b) Confirm your answer to part (a) by graphing the function.
Solution:
76 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 44. (a) The denominator of = 2+ 1
3 − 22 = 2+ 1
(3 − 2)is equal to zero when
= 0and = 32(and the numerator is not), so = 0 and = 15 are vertical asymptotes of the function.
(b)
45. (a) () = 1
3− 1.
From these calculations, it seems that lim
→1− () = −∞ and lim
→1+ () = ∞.
()
05 −114
09 −369
099 −337
0999 −3337 09999 −33337 099999 −33,3337
()
15 042
11 302
101 330
1001 3330 10001 33330 100001 33,3333 (b) If is slightly smaller than 1, then 3− 1 will be a negative number close to 0, and the reciprocal of 3− 1, that is, (),
will be a negative number with large absolute value. So lim
→1− () = −∞.
If is slightly larger than 1, then 3− 1 will be a small positive number, and its reciprocal, (), will be a large positive number. So lim
→1+ () = ∞.
(c) It appears from the graph of that lim
→1− () = −∞ and lim
→1+ () = ∞.
46. (a) From the graphs, it seems that lim
→0
tan 4
= 4. (b)
()
±01 4227 932
±001 4002 135
±0001 4000 021
±00001 4000 000
47. (a) Let () = (1 + )1.
()
−0001 271964
−00001 271842
−000001 271830
−0000001 271828 0000001 271828 000001 271827 00001 271815 0001 271692
It appears that lim
→0(1 + )1≈ 271828, which is approximately .
In Section 3.6 we will see that the value of the limit is exactly .
(b)
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2
