Use l'H^opital's rule to find the limit.
-$\lim _ { x \rightarrow 0 } \frac { \sin 4 x } { \tan 3 x }$
A) $- \frac { 4 } { 3 }$
B) $\frac { 4 } { 3 }$
C) 0
D) $\frac { 3 } { 4 }$

Question

Quizplus · Accepted Answer

Using l'Hopital's rule, 
$$\lim_{x\rightarrow 0} \frac{\sin 4x}{\tan 3x}=\lim_{x\rightarrow 0} \frac{4\cos 4x}{3\sec^2 3x}=\frac{4}{3}\lim_{x\rightarrow 0} \frac{\cos 4x}{\sec^2 3x}=\frac{4}{3}\cos 0\cdot \frac{1}{\sec^2 0}=\frac{4}{3}\cdot 1=\frac{4}{3}$$

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Use L'H^opital's Rule to Find the Limit $\lim _ { x \rightarrow 0 } \frac { \sin 4 x } { \tan 3 x }$