Question

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

• A. $0$
• B. $\frac{1}{2}$
• C. $2$
• D. None of these

Answer 1

The correct option is B $\frac{1}{2}$

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