By applying L'Hopital's rule twice,
[tex]\displaystyle \lim_{x\to\infty} \frac{\log^2(x)}{x} = \lim_{x\to\infty} \frac{\frac{2\log(x)}x}1 \\\\ = \lim_{x\to\infty} \frac{2\log(x)}x \\\\ = \lim_{x\to\infty} \frac{\frac2x}1 \\\\ = \lim_{x\to\infty}\frac2x = \boxed{0}[/tex]
