#Made Easy

Question

 

Find the value of $\lim_{x \rightarrow 0 } \dfrac{x \tan2x - 2x\tan x}{(1-\cos 2x)^2} \rule{1 in}{.5 pt}.$ 

Comments

$\frac{x\tan2x -2x\tan x}{(1- \cos2x)^2} = \frac{x\tan2x -2x\tan x}{(1- (1 – 2\sin^2x))^2} = \frac{x\tan2x -2x\tan x}{4\sin^4x}$

$= \frac{\frac{2x \tan x}{1 \ – \tan^2 x } \ – 2x \tan x}{4\sin^4x} = \frac{2x\tan x}{4 \sin^4x}\frac{\tan^2 x}{1 \ – \ \tan^2 x}$

$= \frac{1}{2\frac{\sin x}{x}} \frac{1}{\cos^3 x} \frac{1}{1 \ – \ \tan^2 x}$

Now, when $x \rightarrow 0$ then $\frac{\sin x}{x} =1, \cos^3x = 1$ and $1 \ – \ \tan^2 x = 1$

So, Answer = $\frac{1}{2}$

Answer 1

The solution of the Limit is given below: