Question

$\underset{x\to 0}{\lim }\frac{x\tan 2x-2x\tan x}{{\left(1-\cos 2x\right)}^2}$ equals :

• A. $\frac{1}{4}$
• B. 1
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (C)

Given:$\underset{x\to 0}{\lim }\frac{x\tan 2x-2x\tan x}{(1-\cos 2x{)}^2}$
$\Rightarrow \underset{x\to 0}{\lim }\frac{\frac{2x\tan x}{1-{\tan }^{2}x}-2x\tan x}{{\left(1-1+2{\sin }^{2}x\right)}^2}\Rightarrow \underset{x\to 0}{\lim }\frac{2x\tan x}{1-{\tan }^{2}x}\left(\frac{1-1+{\tan }^{2}x}{4{\sin }^{4}x}\right)$
$\Rightarrow \underset{x\to 0}{\lim }\frac{1}{2}\left(\frac{x}{\sin x}\right)\left(\frac{{\tan }^{3}x}{x^3}\right)\left(\frac{x^3}{{\sin }^{3}x}\right)=\frac{1}{2}$