$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}$

solution:

Given $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{2}}$

We know that $\sin 2 x=2 \sin x \cos x$, using this formula we get

Again by taking $2 \sin x$ common we get

$\lim _{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^{3}}$

Now we have $\cos x=1-2 \sin ^{2}(x / 2)$

Using this identity above equation can be written as

$\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^{3}}=\lim _{x \rightarrow 0} \frac{2 \sin x\left(1-\left(1-2 \sin ^{2}\left(\frac{x}{2}\right)\right)\right.}{x^{3}}$

On simplifying we get

$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin x\left(1-\left(1-2 \sin ^{2}\left(\frac{x}{2}\right)\right)\right.}{x^{3}} \lim _{x \rightarrow 0} \frac{4 \sin x \sin ^{2}\left(\frac{x}{2}\right)}{x^{3}}$