5 Steps to a 5 AP Calculus AB 2019 - William Ma

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

.