Question

${lim}_{x\to 0}{\log }_{\left({\tan }^{2}x\right)}\left({\tan }^{2}2x\right)=$

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

Answer 1

The correct option is A $1$

$\underset{x\to 0}{lim}{\log }_{{\tan }^{2}x}\left({\tan }^{2}2x\right)$

Let,

$\begin{array}{c}y={\log }_{{\tan }^{2}x}{\tan }^{2}2x\\ =\log {\tan }^{2}2x={\left({\tan }^{2}x\right)}^y\\ =2\log \tan 2x=2y\log \tan x\\ \Rightarrow y=\frac{\log \tan 2x}{\log \tan x}\\ \Rightarrow {lim}_{x\to 0}y={lim}_{x\to 0}\frac{\log \tan 2x}{\log \tan x}\end{array}$

Using L-Hospital Rule

$\begin{array}{c}\Rightarrow {lim}_{x\to 0}y={lim}_{x\to 0}\frac{\frac{1}{\tan 2x}\times {\sec }^{2}2x\times 2}{\frac{1}{\tan x}{\sec }^{2}x}\\ \Rightarrow {lim}_{x\to 0}y={lim}_{x\to 0}\frac{2{\sec }^{2}2x}{\tan 2x}\times \frac{\tan x}{{\sec }^{2}x}\\ \Rightarrow {lim}_{x\to 0}y={lim}_{x\to 0}\frac{2\tan x}{\tan 2x}\left({lim}_{x\to 0}\sec x=4\right)\end{array}$

$\begin{array}{c}\Rightarrow {lim}_{x\to 0}y={lim}_{x\to 0}\frac{2\frac{\tan x}{x}\times x}{\frac{\tan 2x}{2x}\times 2x}\\ \Rightarrow {lim}_{x\to 0}y={lim}_{x\to 0}1\left({lim}_{x\to 0}\frac{\tan x}{x}=1\right)\\ \Rightarrow {lim}_{x\to 0}{\log }_{{\tan }^{2}x}\left({\tan }^{2}2x\right)=1\end{array}$