Question

$\underset{x\to 0}{lim}\frac{x}{2\sin x-\sin 2x}$

Answer 1

We have,

${lim}_{x\to 0}\left(\frac{x}{2\sin x-\sin 2x}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to 0}\left(\frac{1}{2\cos x-2\cos 2x}\right)$

$=\left(\frac{1}{2\cos 0-2\cos 0}\right)$

$=\frac{1}{2-2}$

$=\frac{1}{0}$

$=\infty$

Hence, this is the answer.