Answer
$0$
Work Step by Step
Here, we have $\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{(\ln x)^2}{1/x^2}$
But $\lim\limits_{x \to 0^{+}} \dfrac{2 \ln x/x}{-1/x^2}=\dfrac{-\infty}{-\infty}$
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{\frac{2}{ x}}{\frac{1}{x^2}}=(2) \cdot (0)=0 $
