Answer
$0$
Work Step by Step
When $\lim\limits_{x \to \infty} f(x)$ exists, then the sequence $a_n=f(n)$ converges to the same limit, so we have:
$ \lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} f(x)$
Now, $ \lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{(\ln x )^2}{x} \\=\lim\limits_{x \to \infty} \dfrac{\sin (\pi/x)}{1/x}$
We see that this limit shows the indefinite form of $\dfrac{\infty}{\infty}$, so we apply L-Hopital's rule.
Thus, $ \lim\limits_{n \to \infty} a_n =2 \lim\limits_{x \to \infty} \dfrac{\ln x }{x} \\=\lim\limits_{x \to \infty}\dfrac{1}{x} \\=0 $
