Question

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

Answer 1

$\begin{array}{c}Wehave\\ {lim}_{x\to 0}\frac{1}{x^2}-\cot x\\ =\frac{\tan x-x^2}{x^2\tan x}\\ =\frac{{\sec }^{2}x-2x}{2x\tan x-x^2{\sec }^{2}x}\\ =\frac{1-0}{0+0}\\ =\frac{1}{0}\\ =\infty (Notdefined.)\end{array}$