Question

$\displaystyle\lim_{x\to0} \frac{x^{2} \left(\tan2x -2\tan x\right)^{2}}{\left(1-\cos2x\right)^{4}}= $

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

Answer 1

Answer: (D)

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