Question

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

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

Answer 1

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

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

$=\underset{x\to 0}{lt}\frac{x[\frac{2\tan x}{1-{\tan }^{2}x}-2\tan x]}{(2{\sin }^{2}x{)}^2}$

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

$=\underset{x\to 0}{lt}\frac{\frac{x\tan x}{1-{\tan }^{2}x}[x-1+{\tan }^{2}x]}{2{\sin }^{4}x}$

$=\underset{x\to 0}{lt}\frac{x{\tan }^{3}x\left(\frac{{\sin }^{3}x}{{\cos }^{3}x}\right)/x^4}{(1-{\tan }^{2}x).2{\sin }^{4}x/xy}$

$=\underset{x\to 0}{lt}\frac{{\left(\frac{\sin x}{x}\right)}^3.\frac{1}{{\cos }^{3}x}}{2{\left(\frac{\sin x}{x}\right)}^4(1-{\tan }^{2}x)}=\frac{1.1}{2.1(1-0)}=\frac{1}{2}$

Given $\underset{x\to 0}{lt}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x{)}^2}=\frac{1}{2}$