If $f(x)$ is the integral of $\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0$. Find $\lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right) ; }$,& ;where $f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}$

$f(x)=\dfrac{2\sin x-\sin 2x}{x^3}$ $x\neq 0$

$f'(x)=\dfrac{df(x)}{dx}=\dfrac{2\sin x-\sin 2x}{x^3}$

$=\dfrac{2\sin x}{x}\left(\dfrac{1-\cos x}{x^2}\right)$

$\Rightarrow \displaystyle \lim_{x\rightarrow 0}{f'(x)}=\displaystyle \lim_{x\rightarrow 0}{2\left(\dfrac{\sin x}{x}\right)}\left( \dfrac{2\sin^2(x/2)}{x^2}\right)$

$\Rightarrow \displaystyle 4\cdot 1 \lim_{x\rightarrow 0}{\left(\dfrac{\sin^2(x/2)}{4\times (x/2)^2}\right)}$

$=1$