Question

Evaluate the limit: $\underset{x\to 0}{lim}\frac{1-\cos 2x}{3{\tan }^{2}x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{1-\cos 2x}{3{\tan }^{2}x}$

It becomes $\frac{0}{0}$ form.

$\underset{x\to 0}{lim}\frac{1-\cos 2x}{3{\tan }^{2}x}[\because 1-\cos 2x=2{\sin }^{2}x]$

Dividing numerator and denominator by $x^2,$
we get

$=\underset{x\to 0}{lim}\frac{\frac{2{\sin }^{2}x}{x^2}}{\frac{3{\tan }^{2}x}{x^2}}$

$=\underset{x\to 0}{lim}\frac{2\times \frac{\sin x}{x}\times \frac{\sin x}{x}}{3\times \frac{\tan x}{x}\times \frac{\tan x}{x}}$

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

$=\frac{2\times 1\times 1}{3\times 1\times 1}$

$=\frac{2}{3}$