Question

Solve:
$\underset{x\to 0}{lim}x\log \sin x$

Answer 1

$\underset{x\to 0}{lim}xlog\left(sinx\right)=\underset{x\to 0}{lim}\frac{log\left(\sin x\right)}{\frac{1}{x}}$

and we can see that when we will put $x=0$ it is $\frac{\infty }{\infty }$ form.

So,

using L hospital rule,

$=\underset{x\to 0}{lim}\frac{\frac{d}{dx}log\left(\sin x\right)}{\frac{d}{dx}\left(\frac{1}{x}\right)}$

$=\underset{x\to 0}{lim}\frac{\frac{1}{\sin x}\times \frac{d}{dx}\left(\sin x\right)}{-\frac{1}{x^2}}$

$=\underset{x\to 0}{lim}-x^2\cot x$

$=0$

As when x tends to $0$ $\cot x$ tends to infinity but it is multiplied by zero.

So, Answer is $0$.