Question

The value of ${lim}_{x\to 0}(\frac{1}{x^2}-\cot x)$ equals ?

• A. $1$
• B. $0$
• C. $\infty$
• D. $Doesnotexist$

Answer 1

The correct option is C $\infty$

$li{m}_{x\to 0}(\frac{1}{x^2}-\frac{\cos x}{\sin x})=\left[\frac{\sin x-x^2\cos x}{x^2\sin x}\right]=\frac{0}{0}$

Use L - hospital rule.

$li{m}_{x\to 0}\frac{\cos x-(2x.\cos x-\sin x.x^2)}{2x\sin x+\cos x.x^2}$

$=\frac{1-0+0}{0+0}=\frac{1}{0}=\infty$

Value $=\infty$