Question

Solve:
$\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}$

Answer 1

$\begin{array}{c}{lim}_{x\to 0}\frac{2\sin x-\sin 2x}{x^3}\\ {lim}_{x\to 0}\frac{2\sin x-2\sin x.\cos x}{x^3}\\ {lim}_{x\to 0}\frac{2\sin x\left(1-\cos x\right)}{x^3}\\ {lim}_{x\to 0}\frac{2\sin x}{x}\left(\frac{1-\cos x}{x^2}\right)\\ 2{lim}_{x\to 0}\frac{\sin x}{x}.{lim}_{x\to 0}\left(\frac{1-\cos x}{x^2}\right)\\ 2\times 1\times \left(\frac{-1}{2}\right)=-1\end{array}$