# ) urn ```1_
y
=
4, then lim(x
2 -1-
x
=
5x]
=
1. lim(2x + 1)
2.
g(x)]
=
iimf(x) — iiing(x)
iirnf(x) -I- (—1)liing(x)
urn f(x) -I- iim(—1)g(x)
urn [f(x) -+ (—1)g(x)]
Squeeze Theorem
2
such that
0
&lt;
Is
—
,6
1
inin{6
,
2
Is
cI
&lt;6
L.
L
n
63
—
&lt;
f(x)
&lt;f(x)
:
g(x)
g(x)
‘
=
......_..
x—0
(x)
2
f
—2, then lirn—
x—c g(r)
=
=
1
=
lim (3x
2
—
1)
—
X
4. If lirnf(x)
x-’O
=
x—2
5 —
3x
2x+1
lim [(2x
2+
5 1mm
4.
1)(7x2 -I-
3. lirn [(2x &plusmn; 1)(x — 3)]
x—0
I
6.
13)]
Land tim g(x)
=
=
sinx
1mm—
iim[f(x)-Lg(x)
x—O
=
X
1 for
C,
1. It follows
—-3 7 — 2,2
4x-i- 1
L, then
lim
=
I
/6 g(s) = (sin x)/x, and h(s)
2
x
him h(s) and so, by Theorem B,
SOLUTION Let f(x)
that lini f(x)
r-O
(sin x)/x
h(x) &lt; L + e
h(x)
Assume that we have proved 1 — 5216
=
-
cI
all x near but different from 0. What can we conclude about urn
• EXAMPLE 9
—
63}. Then
0&lt;
L—e&lt;h(x)&lt;L-+c
2
0&lt;lx—cI&lt;6
We conclude that lim g(x)
Let6
Choose 63 so that
and
0&lt;lx—cI&lt;i=L—s&lt;f(x)&lt;L-+c
Proof (Optional) Let c &gt; 0 be given. Choose 6 such that
3)f(x)
=
—
g(r)
h(.x) for all x near
Let f, g, and h be functions satisfying f(.x)
except possibly at c.If iirnf(x) = liinh(x) = L, then linlg(.x) = L.
Theorem B
In Problems 1—12, use Theorem A to find each of the limits.
Justify each step by appealing to a numbered statement, as in
Examples 1—4.
+
x—’c
4 and tim g(x)
Problem Set 1.3
x—c
3. If limf(x)
[f(s)
The Squeeze Theorem You have likely heard someone say, “I was caught
between a rock and a hard place.” This is what happens to g in the following
theorem (see Figure 2),
lim
Proof of Statement 5
g(x)=_2,thenh
2
lfih
V
g2(x) + 12=
2. 9
1. If limf(x)
Concepts Review
C
Chapter 1 Limits
Figure 2
L
72
—
3 &plusmn;
9w
19)_h/2
10.
8.
-i-2x
2
V5x
urn \/—3w -)- 7w
2
urn
urn
w--2
-)--4w+4
2
w
iiin[If(t)I
-I-
13g(t)I]
=
=
5
1
—
32
34. f(s)
=
=
=
=
jg(x)IIf(x)
—
LI
—
&lt;
Is
—
cI
&lt;6
=
Ig(x)I
&lt;
Ml
+
1
M, then there is a number 5 such
=
0
L
cl.
=
=
[f(x)
=
0.
=
0.
g(x)
exists;
X
4 + 4x
+ x
3
48.
urn [x
2 + 2x]
46. hm (x — [x]])
x-i-
urn
x__ir+
44. urn
42.
=
0.
perimeter of R
perimeter of Q
Answers to Concepts Review: 1. 48 2. 4
3. —8;—4+5c 4.0
51. Let y = V and consider the points M, N, 0, and P with
coordinates (1,0), (0,1), (0,0), and (x, y) on the graph of y = V,
respectively. Calculate
perimeter of tSNOP
area of ANOP
(a) jim
jim
x—o perimeter of t\MOP (b) x—O
area of 1M0P
lirn
x—O
50. Let R be the rectangle joining the midpoints of the sides
of the quadrilateral Q having vertices (&plusmn;x, 0) and (0, &plusmn;1).
Calculate
49. Suppose that f(x)g(x) = 1 for all x and urn g(x)
x a
Prove that lim f(x) does not exist.
IxI
(3x — 1) 2
(x2+1)x]
‘—3
47• jim -x—O
x—2
}+
hm
x_*_3*
45 jim
43.
41.
In Problems 41—48,find each of the right-hand and left-hand limits
or state that they do not exist.
iimf(x) or urn g(x) exists.
(b) jim [f(x) g(x)] exists, this does not imply that either
.
limlf(x)I
iim[f(x) — LI
+ g(x)] exists, this does not imply that either
iimf(x) or urn
(a) urn
40. Find examples to show that if
39. Provethatlimlxi
38. Provethatiimf(x)
37. Provethatiirnf(x)
36. Prove Statement 7 of Theorem A by first giving an c—b
proof that urn [1/g(x)] = i/[lim g(x)] and then applying State
ment 6.
0
Now show that if urn g(x)
that
73
Theorem B of the previous section says that limits of polynomial functions can
always be found by substitution, and limits of rational functions can be found by
substitution as long as the denominator is not zero at the limit point. This substitu
tion rule applies to the trigonometric functions as well. This result is stated next.
-f-
Mu
ILIIg(x) MI
—
— Lg(x) -f- Lg(x) — LMI
Ig(x)[f(x) — L] + L[g(x)
If(x)g(x)
1.4
Limits Involving
— LMI
Trigonometric Functions
If(x)g(x)
x
—
3
2 -I- 2x + 1
3x
f(2)1/(x — 2) for each
32. f(s)
—
35. Prove Statement 6 of Theorem A. ifint:
33. f(x)
31. f(x)
3]4
30. lim[f(u) + 3g(u)j
3
28. hrn[f(i)
In Problems 31—34, find urn [f(x)
given functionf
29.
27. IiinY[f(x) +
—
in Problems 25—3U, find the limits if Jim f(s) = 3 and
urn g(x) = —1 (see Example 4).
2f(x)—3g(x)
25. limn\/f2(x) -I- g
(x)
2
26. lien
x—a f(x) -I-- g(x)
24.
—
(w + 2)(w
2 — w — 6)
—
+ 1
+
—
-1- 6
-x
2
1-x—2
— 14x
51
20. lini
x—’—3 x
—1
2
x
2 — 4x — 21
ux&plusmn;2u2x
2
u
+uxxu
2
,x
22. urn
21. urn
u——2
x—1
2
u
u 6
2 + 2x — 3
x
-I2 — 6x1r 4’ir
2x
23. Jim
x—’l
19. Urn
—
2
5x
x—2
—
x—-1 2
16. jim
x—2
14. urn
6x + lix — 6
2
17. jim
x——1 x
3 + 4x
2 — 19i —I- 14
2 + 7x + 10
x
18. liin
2
xx+2
x—1
15. jim
— 2x —3
x+1
2 X
x—
2 -I- 4
13. Jim
in Problems 13—24, find the indicated limit or state that it does not
exist. In many cases, you will want to do some algebra before
trying to evaluate the limit.
5
w-.
4
12. lim(2w
)
15)13
Iim”’_i
11. y2
y&plusmn;4
3 -I9. urn (2t
7. ijrnV’
Section 1.4 Limits Involving Trigonometric Functions
```