Question

The value of $\underset{x\to 1}{lim}{\left(\tan \frac{x\pi }{4}\right)}^{\tan \frac{\pi x}{2}}$ is

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

Answer 1

The correct option is B $e^{-1}$

$\underset{x\to 1}{lim}{\left(\tan \frac{x\pi }{4}\right)}^{\tan \frac{\pi x}{2}}\left(1^{\infty }from\right)$
$e^{\underset{x\to 1}{lim}}(\tan \frac{x\pi }{4}-1)\tan \frac{\pi x}{2}$
$=e^{\underset{x\to 1}{lim}}\left(\frac{\sin \frac{\pi x}{4}-\cos \frac{\pi x}{4}}{\cos \frac{\pi x}{4}}\right)\frac{2\sin \frac{\pi x}{4}\cos \frac{\pi x}{4}}{\cos \frac{\pi x}{2}}$
$=e^{\underset{x\to 1}{lim}}\frac{2{\sin }^2\frac{\pi x}{4}-2\sin \frac{\pi x}{4}\cos \frac{\pi x}{4}}{\cos \frac{\pi x}{2}}$
$=e^{\underset{x\to 1}{lim}}\left(\frac{1-\cos \frac{\pi x}{2}-\sin \frac{\pi x}{2}}{\cos \frac{\pi x}{2}}\right)$
$=e^{\underset{x\to 1}{lim}}(\frac{1-\sin \frac{\pi x}{2}}{\cos \frac{\pi x}{2}}-1)$
$=e^{\underset{x\to 1}{lim}}(\frac{\cos \frac{\pi x}{2}}{1+\sin \frac{\pi x}{2}}-1)$
= $e^{-1}$