Consider the graph of the function and approximate $\lim _ { x \rightarrow 0 } \frac { \sin ( - 8 x ) } { x }$, if it exists.

A) $- 1$
B) $- 4$
C) $- 8$
D) 1
E) The limit does not exist.

Question

Consider the graph of the function and approximate $\lim _ { x \rightarrow 0 } \frac { \sin ( - 8 x ) } { x }$, if it exists.

A) $- 1$ 
B) $- 4$ 
C) $- 8$ 
D) 1 
E) The limit does not exist.

Quizplus · Accepted Answer

From the graph of the function, we can see that as $x$ approaches $0$, $\sin(-8x)$ approaches $0$ and $x$ approaches $0$ from the right side (since the graph only shows the function for $x \geq 0$). Therefore, we have the form $\frac{0}{0}$ when we try to evaluate the limit, so we can apply L'Hopital's rule:

\begin{align*}
\lim_{x 	o 0} \frac{\sin(-8x)}{x} &= \lim_{x 	o 0} \frac{-8\cos(-8x)}{1} \
&= -8\cos(0) \
&= \boxed{-8}.
\end{align*}

Consider the Graph of the Function and Approximate lim⁡x→0sin⁡(−8x)x\lim _ { x \rightarrow 0 } \frac { \sin ( - 8 x ) } { x }limx→0​xsin(−8x)​

Consider the Graph of the Function and Approximate $\lim _ { x \rightarrow 0 } \frac { \sin ( - 8 x ) } { x }$