MA 3972-MA-Book May 7, 2018 9:52
Limits and Continuity 95
Example 2
Evaluate the limit: limx→−∞
3 x− 10
4 x^3 + 5
.
Divide every term in the numerator and denominator by the highest power ofx. In this
case, it isx^3. Thus, limx→−∞
3 x− 10
4 x^3 + 5
=xlim→−∞
3
x^2
−
10
x^3
4 +
5
x^3
=
0 − 0
4 + 0
= 0.
Verify your result with a calculator. (See Figure 6.2-4.)
[−4, 4] by [−20, 10]
Figure 6.2-4
Example 3
Evaluate the limit: limx→∞
1 −x^2
10 x+ 7
.
Divide every term in the numerator and denominator by the highest power ofx. In this
case, it isx^2. Therefore, limx→∞
1 −x^2
10 x+ 7
=xlim→∞
1
x^2
− 1
10
x
+
7
x^2
=
xlim→∞
(
1
x^2
)
−xlim→∞(1)
xlim→∞
(
10
x
)
+xlim→∞
7
x^2
.The limit
of the numerator is−1 and the limit of the denominator is 0. Thus, limx→∞
1 −x^2
10 x+ 7
=−∞.
Verify your result with a calculator. (See Figure 6.2-5.)
[−10, 30] by [−5, 3]
Figure 6.2-5
Example 4
Evaluate the limit: limx→−∞
2 x+ 1
√
x^2 + 3
.
Asx→−∞,x<0 and thus,x=−
√
x^2. Divide the numerator and denominator byx(not
x^2 since the denominator has a square root). Thus, you have limx→−∞
2 x+ 1
√
x^2 + 3
=xlim→−∞
2 x+ 1
√x
x^2 + 3
x