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Compute the Limit [tex]\lim_{x \to \infty} \frac{(log(x))^{2} }{x}[/tex]

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LammettHashidn

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]

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