Question

Evaluate $\underset{x\to 1}{lim}{\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}$

Answer 1

${lim}_{x\to 1}{\left(\tan \frac{\pi x}{4}\right)}^{\tan \frac{\pi x}{2}}=l$ (say) [is of the form $1^{\infty }$]

$l={lim}_{x\to }{[1+\tan \frac{\pi x}{4}-1]}^{\tan \frac{\pi x}{2}}$

$l=e^{{lim}_{x\to }\left[\tan \frac{\pi x}{4}-1\right]/\tan \frac{\pi x}{2}}$ [Because ${lim}_{x\to a}{\left[f\left(x\right)\right]}^{\left[g\left(x\right)\right]}=e^{{lim}_{x\to a}[f\left(x\right)-1]/g\left(x\right)}$} {when, ${lim}_{x\to a}f\left(x\right)=1\&{lim}_{x\to a}g\left(x\right)=\infty$}

$l=e^{{lim}_{x\to }\left[\frac{\left({\sec }^2\frac{\pi }{4}x\right).\frac{\pi }{4}}{\left({\sec }^2\frac{\pi }{2}x\right).\frac{\pi }{2}}\right]}$ [L' Hospital Rule]

$l=e^{\left[4/\infty \right]}$

$l=e^0$

$l=1$