Use l'H^opital's rule to find the limit.
-$\lim _ { x \rightarrow \theta } \frac { \sin 9 x } { 5 x }$
A) $\frac { 9 } { 5 }$
B) 1
C) 0
D) $\frac { 1 } { 5 }$

Question

Quizplus · Accepted Answer

Applying l'Hopital's rule, we get:
$$\lim_{xightarrow 	heta}\frac{\sin 9x}{5x}=\lim_{xightarrow 	heta}\frac{9\cos 9x}{5}= \frac{9\cos 9	heta}{5} = \frac{9}{5}\cos 9	heta$$
Since $\cos$ is bounded between $-1$ and $1$, we have $\frac{9}{5}\cos 9	heta$ is also bounded and the limit exists. Thus, the answer is $\frac{9}{5}$.

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