Question

${lim}_{x\to 0}\frac{1-co{s}^{3}x}{xsinxcotx}$

Answer 1

We have,

${lim}_{x\to 0}\frac{1-{\cos }^{3}x}{x\sin x\cot x}$

$={lim}_{x\to 0}\frac{1-{\cos }^{3}x}{x\sin x\frac{\cos x}{\sin x}}$

$={lim}_{x\to 0}\frac{1-{\cos }^{3}x}{x\cos x}\left(\frac{0}{0}form\right)$

$\text{On}\text{applying}\text{l}{}^{\prime }\text{Hospital}\text{rule}$

$={lim}_{x\to 0}\frac{0-3{\cos }^{2}x(-\sin x)}{x\frac{d}{dx}\left(\cos x\right)+\cos x\frac{dx}{dx}}$

$={lim}_{x\to 0}\frac{3{\cos }^{2}x\sin x}{x(-\sin x)+\cos x}$

$\text{Taking limit}\text{and}\text{we}\text{get,}$

$=\frac{3\sin 0{\cos }^{2}0}{0(-\sin 0)+\cos 0}$

$=\frac{0}{1}$

$=0$

Hence, this is the answer.