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