Question

$\displaystyle\lim_{x \to0} \frac{x \tan2x - 2x \tan x}{\left(1-\cos2x\right)^{2}} $ equals :

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

Answer 1

Answer: (C)

Given:$ \displaystyle\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}} $
$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \frac{\frac{2 x \tan x}{1-\tan ^{2} x}-2 x \tan x}{\left(1-1+2 \sin ^{2} x\right)^{2}} \Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{2 x \tan x}{1-\tan ^{2} x}\left(\frac{1-1+\tan ^{2} x}{4 \sin ^{4} x}\right) $
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \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} $