Answer
$e^{1}$
Work Step by Step
Here, we have $\ln f(x)=\ln (\dfrac{\ln x}{x-1}) \implies f(x)=e^{( \frac{\ln x}{x-1})}$
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 limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 1} e^{(\lim\limits_{x \to 1} \frac{1/x}{1})}=e^{( \frac{1/1}{1})}=e^{1}$
