Question

Solve:
$\underset{x\to 0}{lim}\frac{e^{\sin 3x}-1}{\log \left(1+\tan 2x\right)}=\frac{3}{2}$

Answer 1

${lim}_{x\to 0}\frac{e^{\sin 3x}-1}{\log (1+\tan 2x)}$

use L'H rule

$={lim}_{x\to 0}\frac{e^{\sin 3x}\cdot \cos 3x\cdot 3}{\frac{1x{\sec }^2\left(2x\right)\cdot 2}{(1+\tan 2x)}\cdot 2}$

$=\frac{3}{2}$