$\text { Evaluate the limit } \lim _ { x \rightarrow 10 } \frac { - 9 ( x - 10 ) } { x ^ { 2 } - 100 } \text { first by using techniques from Chapter } 1 \text { then }$ by using L'Hopital's Rule.
A) $\frac { 9 } { 20 }$
B) $- \frac { 9 } { 100 }$
C) $- \frac { 9 } { 20 }$
D) 0
E) does not exist

Question

$\text { Evaluate the limit } \lim _ { x \rightarrow 10 } \frac { - 9 ( x - 10 ) } { x ^ { 2 } - 100 } \text { first by using techniques from Chapter } 1 \text { then }$ by using L'Hopital's Rule. 
A) $\frac { 9 } { 20 }$
B) $- \frac { 9 } { 100 }$
C) $- \frac { 9 } { 20 }$
D) 0
E) does not exist

Quizplus · Accepted Answer

Using techniques from Chapter 1:

To evaluate the limit we can simply substitute 10 for x because the limit is in indeterminate form (0/0) so we get,

$$\lim _ { x \rightarrow 10 } \frac { - 9 ( x - 10 ) } { x ^ { 2 } - 100 }=\frac {-9(10-10)}{(10)^2-100}=\frac{-9}{20}$$

Using L'Hopital's Rule:

$$\begin{aligned} \lim _ { x \rightarrow 10 } \frac { - 9 ( x - 10 ) } { x ^ { 2 } - 100 }&=\lim _ { x \rightarrow 10 } \frac {-9}{2x} \cdot \frac{1}{x+10} &&\text{(apply L'Hopital's Rule)}\ &=\frac{-9}{2\cdot 10}\cdot \frac{1}{10+10}\ &=-\frac{9}{20} \end{aligned}$$

Since both methods yield the same result, the answer is C, -9/20.

By Using L'Hopital's Rule 920\frac { 9 } { 20 }209​

By Using L'Hopital's Rule $\frac { 9 } { 20 }$