Since $f(x)$ is integral function of $\large\frac{2\sin x-\sin 2x}{x^3}$
$\therefore f'(x)=\large\frac{2\sin x-\sin 2x}{x^3}$
$\Rightarrow \lim\limits_{x\to 0}f'(x)=\lim\limits_{x\to 0}\large\frac{2\sin x-\sin 2x}{x^3}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2\sin x}{x}\big(\large\frac{1-\cos x}{x^2}\big)$
$\Rightarrow 2\times \large\frac{1}{2}$
$\Rightarrow 1$
Hence (c) is the correct answer.