Question

${\lim }_{x\to 0}\frac{x^2-\tan 2x}{\tan x}$

Answer 1

We have,

${lim}_{x\to 0}\frac{x^2-\tan 2x}{\tan x}$

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

So, apply L-hospital rule

${lim}_{x\to 0}\frac{2x-2{\sec }^{2}2x}{{\sec }^{2}x}$

$=\frac{0-2{\sec }^{2}0}{{\sec }^{2}0}$

$=\frac{-2\times 1^2}{1^2}$

$=-2$

Hence, this is the answer.