Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}$

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

$\underset{x\to 0}{lim}\frac{x\tan 2x-2x\tan x}{x^4}\times \frac{x^4}{(2{\sin }^{2}x{)}^2}$

$\underset{x\to 0}{lim}\frac{x\tan 2x-2x\tan x}{x^4}\times \underset{x\to 0}{lim}\frac{x^4}{4{\sin }^{4}x}$

$\underset{x\to 0}{lim}\frac{x(2x+{\left(\frac{2x}{3}\right)}^3)-2x(x+\frac{x^3}{3})}{x^4}\times \underset{x\to 0}{lim}\frac{1}{4}{\left(\frac{x}{\sin x}\right)}^4$

$\underset{x\to 0}{lim}\left(\frac{2x^2+\frac{8}{3}{x}^4-2x^2-\frac{2x^4}{3}}{x^4}\right)\times \frac{1}{4}$

$=\frac{8}{3}-\frac{2}{3}\times \frac{1}{4}$

$=2\times \frac{1}{4}=\frac{1}{2}$.