# Question #f378e

May 5, 2017

$0$

#### Explanation:

Use L'Hôpital's rule.

If ${\lim}_{x \to c} f \frac{x}{g} \left(x\right)$ is one of the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then ${\lim}_{x \to c} f \frac{x}{g} \left(x\right) = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$, where $'$ means the derivative with respect to $x$.

${\lim}_{x \to \infty} {\ln}^{2} \frac{x}{x}$ is of the form $\frac{\infty}{\infty}$. Thus, ${\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \left({\ln}^{2} \left(x\right)\right)}{\frac{d}{\mathrm{dx}} \left(x\right)} = {\lim}_{x \to \infty} \frac{2 \ln \left(x\right)}{x}$.

This is still of the form $\frac{\infty}{\infty}$. Apply L'Hôpital's rule again: ${\lim}_{x \to \infty} \frac{2 \ln \left(x\right)}{x} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \left(2 \ln \left(x\right)\right)}{\frac{d}{\mathrm{dx}} \left(x\right)} = {\lim}_{x \to \infty} \frac{2}{x} = 0$.

This graph demonstrates that ${\ln}^{2} \frac{x}{x}$ does approach $0$ as $x$ approaches $\infty$ (very slowly).

graph{ln(x)^2/x [-8.89, 8.89, -4.446, 4.446]}