Question

Evaluate:

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

Answer 1

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

$={lim}_{x\to 0}\frac{\left(\frac{\sin 2x+3x}{2x}\right)2x}{\frac{(2x+\tan 3x)}{3x}\times 3x}$

$={lim}_{x\to 0}\frac{(\frac{\sin 2x}{2x}+\frac{3x}{2x})\times 2x}{(\frac{2x}{3x}+\frac{\tan 3x}{3x})\times 3x}$

$=\left(\frac{1+\frac{3}{2}}{\frac{2}{3}+1}\right)\times \frac{2}{3}$ $\left[\begin{array}{c}\because {lim}_{x\to 0}\frac{\sin x}{x}=1,\\ {lim}_{x\to 0}\frac{\tan x}{x}=1\end{array}\right]$

$=\frac{3}{2}\times \frac{2}{3}$

$=1$