Answer
$e$
Work Step by Step
L-Hospital's rule defined as $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
Consider $\ln f(x)=(\ln x) ^{1/\ln x}$
and $\ln f(x)= \dfrac{\ln x}{\ln x}$
or, $ f(x)=1$
Thus, we get $e^{[\lim\limits_{x \to 0^{+}} (1)]}=e^{1}=e$
