Answer
$e^{1/e}$
Work Step by Step
L'-Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\ln f(x)=\frac{\ln (\ln x)}{x-e} \implies f(x)=e^{\frac{\ln (\ln x)}{x-e}}$
Now, $e^{\lim\limits_{x \to e^{+}} ( \frac{\ln(\ln x)}{x})}=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{x \to e^{+}} ( \frac{1/x ln x}{1})}=e^{1/e}$
