Answer
$-\infty$
Work Step by Step
Here, we have $\lim\limits_{x \to 0} f(0)=\dfrac{\infty}{-\infty}$
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 0} \dfrac{\dfrac{2 \ln x}{x}}{\dfrac{\cos x}{\sin x}}=\lim\limits_{x \to 0} \dfrac{2 (\ln x) \sin x}{x \cos x}$
This implies that
$\lim\limits_{x \to 0} \dfrac{2 (\ln x)}{\cos x}(\lim\limits_{x \to 0} \dfrac{ \sin x}{x})$
or, $\lim\limits_{x \to 0} \dfrac{2 (\ln x)}{\cos x}(1)=\dfrac{2 (-\infty)}{1}=-\infty$
