Question

${lim}_{x\to 0}\frac{x^2\cos x}{1-\cos x}$ is

• A. $\frac{3}{2}$
• B. 2
• C. $-\frac{3}{2}$
• D. 1

Answer 1

The correct option is B 2

Given:${lim}_{x\to 0}\frac{x^2\cos x}{1-\cos x}$

$={lim}_{x\to 0}\frac{x^2\cos x}{2{\sin }^2\frac{x}{2}}$ $[\because 1-\cos x=2{\sin }^2\frac{x}{2}]$

$={lim}_{x\to 0}\frac{{\left(\frac{x}{2}\right)}^2.2\cos x}{{\sin }^2\frac{x}{2}}$

${lim}_{x\to 0}\left(\frac{{\left(\frac{x}{2}\right)}^2}{\sin \frac{x}{2}}\right).2\cos x$
$=2$ $[\because {lim}_{x\to 0}\frac{x}{\sin x}=1]$

Hence, the correct option is (a)