Section 1.5 Limits at Infinity, Infinite Limits
81
SOLUTION We often have a vertical asymptote at a point where the denomi
nator is zero, and in this case we do because
2x
urn
2x
.
and
00
lim
—----—
= —CX)
x—1
x—’i
On the other hand,
urn
x—’
f(x)
Fjire
2r
and soy
Figure 8.
=
=
2x
x
—
1
=
2
urn
x— 1
=2
1/x
—
2x
urn
and
2
--
_,—_cxJx
2 is a horizontal asymptote. The graph of y
I
—
2x/(x
1) is shown in
—
8
(;ollcepts Review
ii
I. [o say that x
/ (x) = L means that
C-
means that
to say that
Give your answers in informal
3. If Iimf(x)
I iiiiae.
4. If lim f(x)
2. To say that lim f(x)
liii 1(x)
ii
=
=
means that
—
to say that
means that
6, then the line
=
tote of the graph of y
=
tote of the graph of y
=
asymp
is a
asymp
then the line
00,
=
is a
f(x).
f(x).
Give your answers in infor
language.
Problem Set 1.5
In l’rohlems 1—42, find (lie limits.
I.
jim
=
x
—
7
x(x
-
2. in
5
4.
(2
—
21.
5)(3
-
x)
6.
divide by \/2x
9.
Iim
---=
-
7TX
8.
)2
2
5x
urn
jim
15
1+ 8x
2
x
+
2
4
20
sin
10. lim
5—*(XJ 02
5
16.
18.
n’ 1
7
In
Iimj
In
5.
—
—
.
cxy2y+2
I-lint: Divide numerator and denomi
—
+
X
0
(1
x”
a
1
+
-
+ a _
x + a
1
+
+ b_ 1
x + b,
a
-
lim—
26.
i
where a
0
—
+
28.
lim
(—‘--_;
29. lim
+
30.
—
9
2n + I
9
+ 3
12
X
lim
lim
“==
l)(x + I)
lim
,
“
x” + b
0
h
x”
1
+
27.
(x
lim
x—’c
0, and n is a natural number.
25.
Hint: Divide numerator and denominator
\/‘x + 3
by x. Note that, for x > 0, \/ + 3/i = \/(x2 + 3)/x
.
2
19.
3 + I
9y
lim
+ 7x
2
a
,2
24.
—
-s
lim
.r—’V’
5
—
(32
31.
33.
-
+ \/2x
.
urn
v-’a (x
jim
x-- 3 x
)(3
—
—
—
3
x)
32.
34.
lim
sin C)
irO
Iirn
v—’( r/2)’ COS C)
v-Xr
20. lim
x+4
and
nator by y
.
2
b()
—
14.
3
Mulliply
1-lint:
1im(\/+2x_•r)
Y’
SO
—
12.
15. 11m
2
17. ii
+
+
\/_5).
—
T0
3\/3
limq
23.
—
—
II. Jim
13.
22.
Hrn—
-
2xIOOx
lim (V
cx)
35.
37.
limXX
x—3
36. jim
urn
38.
x—’3
r—’O
X
•
jim
r
+2x—8
-——-,--—-——----
x—4
Ix
0,