# How do you find the Limit of  (ln x)^2 / (3x ) as x approaches infinity?

Aug 1, 2016

0

#### Explanation:

on inspection you can see that the logarithmic numerator is going to grow more slowly than the denominator

but as its $\frac{\infty}{\infty}$ indeterminate , the easy way to show this is to L'Hopital it

${\lim}_{x \to \infty} {\left(\ln x\right)}^{2} / \left(3 x\right)$

By Lhopital
$= {\lim}_{x \to \infty} \frac{2 \left(\ln x\right) \cdot \frac{1}{x}}{3}$

$= 2 {\lim}_{x \to \infty} \frac{\ln x}{3 x}$

which is still $\frac{\infty}{\infty}$ indeterminate so L'Hopital again

$= 2 {\lim}_{x \to \infty} \frac{\frac{1}{x}}{3}$

$= 2 {\lim}_{x \to \infty} \frac{1}{3 x} = 0$

Aug 1, 2016

${\lim}_{x \to \infty} \frac{{\log}_{e} x}{\sqrt{x}} = 0$

#### Explanation:

${\left({\log}_{e} x\right)}^{2} / \left(3 x\right) = \frac{1}{3} {\left(\frac{{\log}_{e} x}{\sqrt{x}}\right)}^{2}$

Now ${e}^{x}$ is monotonic so

${\lim}_{x \to \infty} \frac{{\log}_{e} x}{\sqrt{x}} \equiv {\lim}_{x \to \infty} \frac{{e}^{{\log}_{e} x}}{e} ^ \left\{\sqrt{x}\right\} = {\lim}_{x \to \infty} \frac{x}{e} ^ \left\{\sqrt{x}\right\} \equiv {\lim}_{y \to \infty} {y}^{2} / {e}^{y} = 0$

so

${\lim}_{x \to \infty} \frac{{\log}_{e} x}{\sqrt{x}} = 0$