Question

Value of $\underset{x\to 0}{lim}\frac{(1-\cos 2x{)}^2}{2x\tan x-x\tan 2x}$ is:

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

Answer 1

The correct option is A $2$

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