Question

Evaluate the limit:

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

Answer 1

We have,

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

It becomes 0/0 form.

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

Dividing numerator and denominator by $x^2$,

we get

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

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


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

$=2$

Therefore,

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