Answer
$e^{1/e}$
Work Step by Step
Here, we have $\ln f(x)=\frac{\ln (\ln x)}{x-e}$
and $f(x)=e^{\frac{\ln (\ln x)}{x-e}}$
But $e^{\lim\limits_{x \to e^{+}} ( \frac{\ln(\ln x)}{x})}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$e^{\lim\limits_{x \to e^{+}} ( \frac{1/x ln x}{1})}=e^{ \frac{1/e ln e}{1})}=e^{1/e}$
