Question

${lim}_{x\to 0}\frac{\tan x-\sin x}{{\sin }^{3}x}$

Answer 1

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

Hence, this is the answer.