Answer
$0$
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, $\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{1/x^2}$
Now, $\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{1/x^2}=\dfrac{-\infty}{\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0^{+}} \dfrac{1/ x}{-2/x^3}=\dfrac{0}{2}=0 $
