Answer
$e$
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, $\ln f(x)=(\ln x) ^{1/\ln x} \implies\ln f(x)= \dfrac{\ln x}{\ln x}$
or, $\ln f(x)=\frac{(\ln x)}{\ln x} \implies f(x)=1$
Now, $e^{\lim\limits_{x \to 0^{+}} (1)}=e^{1}=e$