Question

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

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

Answer 1

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

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