$\lim_{x \to 0} \frac{tan x-sinx}{sin^3x}$

$\displaystyle\lim_{x \to 0} \dfrac{\tan x- \sin x}{\sin^3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{\tan x- \sin x}{\sin^3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{\dfrac{\sin x}{\cos x}- \sin x}{\sin^3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{\dfrac{\sin x}{\cos x}- \sin x}{\sin^3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{\dfrac{\cos x}{\sin x}\left(\dfrac{\sin x}{\cos x}- \sin x\right)}{\dfrac{\cos x}{\sin x}.\sin^3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{1- \cos x}{\cos x\sin^2x}\\ =\displaystyle\lim_{x \to 0} \dfrac{2\sin^2\left(\dfrac{x}{2}\right)^2}{\cos x\sin^2x}\\ =\dfrac{1}{2}\displaystyle\lim_{x \to 0} \dfrac{\left(\dfrac{\sin^2\left(\dfrac{x}{2}\right)}{\dfrac{x}{2}}\right)^2}{\cos x\left(\dfrac{\sin^2x}{x}\right)^2}\\ =\dfrac{1}{2}$