Question

The value of ${lim}_{x\to 0}\left(\frac{1}{x^2-cotx}\right)$ is :

Answer 1

$\underset{x\to 0^{+}}{lim}\frac{1}{x^2-\cot x}=\frac{1}{0-(+\infty )}=\frac{1}{-\infty }=0$

$\underset{x\to 0^{-}}{lim}\frac{1}{x^2-\cot x}=\frac{1}{0-(-\infty )}=\frac{1}{\infty }=0$

As RHL$=$LHL

$\underset{x\to 0}{lim}\frac{1}{x^2-\cot x}=0$.