calculus - Finding $lim_{xto +infty}(frac{x+ln x}{ x-ln x})^{frac{x}{ln x}}$

Find $\lim_{x\to +\infty}(\frac{x+\ln x}{ x-\ln x})^{\frac{x}{\ln x}}$. I tried using l'Hospital rule with the continuity of $e$ function. Also tried using Taylor expansion with no success. What should I do? Thank you.

Answer

\begin{align*}
\lim_{x\to+\infty}\left(\frac{x+\log x}{x-\log x}\right)^{\frac x{\log x}}
=&\lim_{x\to+\infty}\left(1+\frac{2\log x}{x-\log x}\right)^{\frac x{\log x}}\\
=&\lim_{x\to+\infty}\left(1+\frac{2}{\frac{x}{\log x}-1}\right)^{\frac x{\log x}}\\
=&\lim_{x\to+\infty}\left(1+\frac{2}{\frac{x}{\log x}-1}\right)^{\frac x{\log x}-1}\left(1+\frac{2}{\frac{x}{\log x}-1}\right)\\
=&e^2
\end{align*}