Answer
$e^{-1}$
Work Step by Step
Here, we have $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 1} e^{(\lim\limits_{x \to 1} \frac{\ln x}{1-x})}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will again apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 1} e^{\frac{1/x}{-1}}=e^{\frac{1/1}{-1}}=e^{-1}$
