Answer
$\dfrac{3}{2}$
Work Step by Step
Given: $\lim\limits_{x\to0}\dfrac{\tan 3x}{\sin 2x}$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim\limits_{x\to0}\dfrac{\dfrac{d}{dx}(\tan 3x)}{\dfrac{d}{dx}(\sin 2x)}=\lim\limits_{x\to0}\dfrac{3\sec^2(3x)}{2\cos 2x}=\dfrac{3}{2}\dfrac{\sec^2(3\times0)}{\cos(2\times0)}=\dfrac{3}{2}$
