Question

Find ${lim}_{x\to 0}\left\{tan\right(\frac{\pi }{4}+x\left)\frac{1}{x}\right\}$.

• A. $e^5$
• B. $e^3$
• C. $e^{-5}$
• D. $e^2$

Answer 1

The correct option is D $e^2$

Consider the given function

$y={lim}_{x\to 0}\left\{\tan {(\frac{\pi }{4}+x)}^{\frac{1}{x}}\right\}$

Taking log both side and we get,

$\log y={lim}_{x\to 0}\left\{log\tan {(\frac{\pi }{4}+x)}^{\frac{1}{x}}\right\}$

$\Rightarrow \log y={lim}_{x\to 0}\left\{\frac{1}{x}log\tan (\frac{\pi }{4}+x)\right\}$

$\Rightarrow \log y={lim}_{x\to 0}\frac{log\tan (\frac{\pi }{4}+x)}{x}$

Its make $\frac{0}{0}from$

Then,

Using L’ hospital rule and we get,

$\log y={lim}_{x\to 0}\frac{\frac{1}{\tan (\frac{\pi }{4}+x)}\times {\sec }^2(\frac{\pi }{4}+x)}{1}$

$\log y={lim}_{x\to 0}\frac{1}{\tan (\frac{\pi }{4}+x)}\times {\sec }^2(\frac{\pi }{4}+x)$

$\log y={lim}_{x\to 0}\frac{\cos (\frac{\pi }{4}+x)}{\sin (\frac{\pi }{4}+x)}\times \frac{1}{co{s}^2(\frac{\pi }{4}+x)}$

$\log y={lim}_{x\to 0}\frac{1}{\sin (\frac{\pi }{4}+x)cos(\frac{\pi }{4}+x)}$

$\log y={lim}_{x\to 0}\frac{2}{2\sin (\frac{\pi }{4}+x)cos(\frac{\pi }{4}+x)}$

$\log y={lim}_{x\to 0}\frac{2}{\sin 2(\frac{\pi }{4}+x)}$

$\log y={lim}_{x\to 0}\frac{2}{\sin 2(\frac{\pi }{4}+x)}$

$\log y={lim}_{x\to 0}2\csc 2(\frac{\pi }{4}+x)$

$\log y={lim}_{x\to 0}2\csc (\frac{\pi }{2}+2x)$

$\log y={lim}_{x\to 0}2sec2x$

Taking limit and we get,

$\log y=2$

${\log }_{e}y=2$

$y=e^2$

Hence, this is the answer.