Question

Evaluate

$\underset{x\to I}{lim}{\left(tan\frac{\pi x}{4}\right)}^{tan\frac{\pi x}{2}}$

Answer 1

We have,

$Let,$

$y={\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}$

Taking log both side and we get,

${lim}_{x\to 1}\log y=\log {lim}_{x\to 1}{\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}$

$\Rightarrow {lim}_{x\to 1}\log y=\tan \frac{\pi x}{2}\log {lim}_{x\to 1}\left(\tan \frac{\pi x}{4}\right)0\times \infty form$

$\Rightarrow {lim}_{x\to 1}\log y=\frac{\log {lim}_{x\to 1}\left(\tan \frac{\pi x}{4}\right)}{\cot \frac{\pi x}{2}}$

Applying L’ Hospital rule and we get,

${lim}_{x\to 1}\log y=\frac{\frac{\pi }{4}{\sec }^2\frac{\pi x}{4}}{\tan \frac{\pi x}{4}}\times \frac{1}{\left(\frac{-\pi }{2}\right){\csc }^2\frac{\pi x}{2}}$

$=\frac{\frac{\pi }{4}{\left(\sqrt{2}\right)}^2}{1}\times \frac{1}{\frac{-\pi }{2}\times 1}=-1$

${lim}_{x\to 1}\log y=\log {lim}_{x\to 1}{\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}$

$\log {lim}_{x\to 1}{\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}=-1$

$\Rightarrow {lim}_{x\to 1}{\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}=e^{-1}$

$\Rightarrow {lim}_{x\to 1}{\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}=\frac{1}{e}$

Hence, this is the answer.