Question

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

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

Answer 1

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

We can write ${lim}_{x\to 0}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x{)}^2}$ as
$={lim}_{x\to 0}\frac{\frac{2x\tan x}{(1-\tan x)}-2x\tan x}{(2{\sin }^{2}x{)}^2}$
$={lim}_{x\to 0}\frac{2x{\tan }^{3}x}{(1-{\tan }^{2}x)4{\sin }^{4}x}$
$={lim}_{x\to 0}\frac{x}{2\sin x{\cos }^{2}x\cos 2x}$
$={lim}_{x\to 0}\frac{x\sec x}{2(\sin x-{\sin }^{3}x)}$
Substituting gives indeterminate form.
Applying L'Hospital's Rule,
$={lim}_{x\to 0}\frac{2x\sec 2x\tan 2x+\sec x}{2(\cos x-3{\sin }^{2}x\cos x)}$
$=\frac{0+1}{2(1-0)}=\frac{1}{2}$
Hence, option 'C' is correct.