Answer
$-\dfrac{1}{2}$
Work Step by Step
Here, we have $\lim\limits_{x \to 1} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 1} \dfrac{-1+1/x)}{1-(1/x)+\ln x}=\dfrac{0}{0}$
Now, again apply L-Hospital's rule.
$\lim\limits_{x \to 1} \dfrac{-1/x^2}{1+x/x^2}=-\dfrac{1/1}{1+1/1}$
or, $-\dfrac{1}{1+1}=-\dfrac{1}{2}$
