Question

Solve $\underset{x\to 0}{lim}\frac{x^2\cos x}{1-\cos x}$

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

Answer 1

The correct option is A $2$

Consider the given function.

$\underset{x\to 0}{lim}\left(\frac{x^2\cos x}{1-\cos x}\right)$

This is $\frac{0}{0}$ form.

So, apply L-Hospital rule,

$\underset{x\to 0}{lim}\left(\frac{x^2(-\sin x)+2x\cos x}{0-(-\sin x)}\right)$

$\underset{x\to 0}{lim}\left(\frac{-x^2\sin x+2x\cos x}{\sin x}\right)$

This is $\frac{0}{0}$ form.

Again, L-Hospital rule

$\underset{x\to 0}{lim}\left(\frac{-x^2\cos x-2x\sin x+2\cos x+2x(-\sin x)}{\cos x}\right)$

$\underset{x\to 0}{lim}\left(\frac{-x^2\cos x-4x\sin x+2\cos x}{\cos x}\right)$

$=\frac{0-0+2\cos 0}{\cos 0}$

$=2$

Hence, this is the answer.