Question

$\underset{x\to 0}{\lim }\frac{3\sin 2x+2x}{3x+2\tan 3x}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{3\sin \left(2x\right)+2x}{3x+2\tan \left(3x\right)}\right]$

$$\begin{gathered} =\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{3\sin 2x}{x}}+2}{3+{\displaystyle \frac{2\tan 3x}{x}}}\right] \\ =\frac{{\displaystyle 3\underset{x\to 0}{\lim }\left(\frac{\sin 2x}{2x}\right)\times 2}+2}{3+2\underset{x\to 0}{\lim }{\displaystyle \left(\frac{\tan 3x}{x}\right)\times 3}} \\ =\frac{6+2}{3+6} \\ =\frac{8}{9} \end{gathered}$$