Question

Evaluate the limit:

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

Answer 1

We have,

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

It becomes 0/0 form.

On dividing numerator and denominator by $x$

$=\underset{x\to 0}{lim}\frac{x-\frac{\tan 2x}{2x}\times 2}{\frac{\tan x}{x}}[\because \underset{x\to 0}{lim}\frac{\tan x}{x}=1]$

$=\frac{0-1\times 2}{1}$

$=-2$

Therefore,

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