Answer
$-\infty$
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, $\lim\limits_{x \to 0} f(0)=\dfrac{\infty}{-\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$ \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
$\lim\limits_{x \to 0} \dfrac{2 (\ln x)}{\cos x}\lim\limits_{x \to 0} (\lim\limits_{x \to 0} \dfrac{2 (\ln x) \sin x}{x \cos x})=\lim\limits_{x \to 0} \dfrac{2 (\ln x)}{\cos x}(1)$
or, $\dfrac{2 (-\infty)}{\cos 0}=-\infty$
