Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{x\tan x}{1-\cos 2x}$

Answer 1

We have

$\underset{x\to 0}{lim}\frac{x\tan x}{1-\cos 2x}$

It becomes 0/0 form.

$=\underset{x\to 0}{lim}\frac{x\tan x}{2{\sin }^{2}x}[\because 1-cos2x=2si{n}^{2}x]$

Dividing numerator and denominator by $x^2$,

we get

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

$=\underset{x\to 0}{lim}\frac{\frac{\tan x}{x}}{2\times \frac{\sin x}{x}\times \frac{\sin x}{x}}$

$[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1,\underset{x\to 0}{lim}\frac{\tan x}{x}=1]$


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

$=\frac{1}{2}$

Therefore,

$\underset{x\to 0}{lim}\frac{x\tan x}{1-\cos 2x}=\frac{1}{2}$