Question

$\underset{x\to 0}{lim}{\left\{\tan (\frac{\pi }{4}-x)\right\}}^{1/x}$ is equal to

• A. $1$
• B. $e$
• C. $e^2$
• D. $e^{-2}$

Answer 1

The correct option is D $e^{-2}$

${lim}_{x\to 0}{\left\{\tan (\frac{\pi }{4}-x)\right\}}^{1/x}={lim}_{x\to 0}\exp \log {\left\{\tan (\frac{\pi }{4}-x)\right\}}^{1/x}=\exp {lim}_{x\to 0}\frac{\log \left\{\tan (\frac{\pi }{4}-x)\right\}}{x}$
Applying L'Hospital rule
$=\exp {lim}_{x\to 0}\frac{-1}{\sin (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-x)}=e^{-2}$
Hence, option 'D' is correct.