Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $-\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

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

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

By expanding $\tan 2x$ and $\cos 2x$ we get

$\frac{(x\tan 2x-2x\tan x)}{{(1-\cos 2x)}^2}=\frac{x\frac{2\tan x}{1-(\tan x{)}^2}-2x\tan x}{{(1-(1-2{\sin }^{2}x\left)\right)}^2}$

$=\frac{2x\tan x-[2x\tan x-2x{\tan }^{3}x]}{4{\sin }^{4}x\times (1-{\tan }^{2}x)}=\frac{2x{\tan }^{3}x}{4{\sin }^{4}x\times (1-{\tan }^{2}x)}$

$=\frac{2x{\tan }^{3}x}{4{\sin }^{4}x\times \left(\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x}\right)}=\frac{2x\frac{{\sin }^{3}x}{{\cos }^{3}x}}{4{\sin }^{4}x\times \left(\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x}\right)}$

$=\frac{x}{2\sin x\times \left({\cos }^{2}x-{\sin }^{2}x\right)\cos x}$

Now applying the limit $x\to 0$,we get

$L=\underset{x\to 0}{lim}\frac{x}{2\sin x}\times \underset{x\to 0}{lim}\frac{1}{\cos x\left({\cos }^{2}x-{\sin }^{2}x\right)}=\underset{x\to 0}{lim}\frac{x}{2\sin x}\times \underset{x\to 0}{lim}\frac{1}{\cos 0({\cos }^{2}0-{\sin }^{2}0)}=\frac{1}{2}[\because \underset{x\to 0}{lim}\frac{x}{\sin x}=1]$

Hence, option D is correct.