Question

${lim}_{x\to 0}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x{)}^2}$ equal ?

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

Answer 1

The correct option is C $\frac{1}{2}$

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

$=\underset{xto0}{lim}\frac{x\tan \left(2x\right)-2x\tan x}{(2{\sin }^{2}x{)}^2}$

$=\frac{1}{4}\underset{x\to 0}{lim}\frac{x\tan 2x-2x\tan x}{{\sin }^{2}x}$

$=\frac{1}{4}\underset{xto0}{lim}\frac{\tan 2x-2\tan x}{{\sin }^{3}x}$

$=\frac{1}{4}\underset{xto0}{lim}\frac{\frac{2\tan x}{1-\tan x}-2\tan x}{{\sin }^{3}x}$

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

$=\frac{1}{4}\underset{x\to 0}{lim}\frac{2}{1-{\tan }^{2}x}\times \underset{x\to 0}{lim}{\left(\frac{\tan x}{\sin x}\right)}^3$

$=\frac{1}{4}\times \frac{2}{1-0}\times 1$

$=\frac{1}{2}$