Question

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

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

Answer 1

Answer: (D)

Given, $\underset{x\to 0}{\lim }\frac{x^2(\tan 2x-2\tan x{)}^2}{(1-\cos 2x{)}^4}$
$=\underset{x\to 0}{\lim }\frac{x^2\left(2x+\frac{(2x{)}^3}{3}+\frac{2}{15}(2x{)}^5+\dots \right)}{{\left[1-\left(1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\frac{(2x{)}^6}{6!}+\dots \right)\right]}^4}$
$=\underset{x\to 0}{\lim }\frac{4x^8\left[\left(\frac{4}{3}-\frac{1}{3}\right)+\frac{2}{15}\left(16x^2-x^2\right)+\dots \right]}{16x^8{\left[1+\frac{x^2}{3}+\dots \right]}^4}$
$=\frac{4}{16}=\frac{1}{4}$