Question

Solve $\underset{x\to 1}{lim}(1+lnx{)}^{sec\frac{\pi x}{2}}$

Answer 1

$\underset{x\to 1}{lim}(1+lnx{)}^{sec\frac{\pi x}{2}}$

when $x=1$ then $\ln x=0$ and $\sec \frac{\pi x}{2}=\infty$

Thus is Limit is of $1^{\infty }$ form

let $L=\underset{x\to 1}{lim}(1+lnx{)}^{sec\frac{\pi x}{2}}$

Taking log on both sides

$\ln L=\underset{x\to 1}{lim}sec\frac{\pi x}{2}\ln (1+lnx)$

$\ln L=\underset{x\to 1}{lim}\frac{\ln (1+lnx)}{cos\frac{\pi x}{2}}$

L'Hospitals rule

$\ln L=\underset{x\to 1}{lim}\frac{\frac{1}{1+lnx}\times \frac{1}{x}}{-sin\frac{\pi x}{2}\times \frac{\pi }{2}}$

$\ln L=\frac{\frac{1}{1+ln1}\times \frac{1}{1}}{-sin\frac{\pi \left(1\right)}{2}\times \frac{\pi }{2}}=-\frac{2}{\pi }$

$L=e^{-\frac{2}{\pi }}$