# How do you find the limit of lnx/x as x->oo?

Jan 16, 2017

${\lim}_{x \to \infty} \ln \frac{x}{x} = 0$

#### Explanation:

As we have:

${\lim}_{x \to \infty} \ln x = + \infty$
${\lim}_{x \to \infty} x = + \infty$

The limit:

${\lim}_{x \to \infty} \ln \frac{x}{x}$

presents itself in the indeterminate form $\frac{\infty}{\infty}$ and we can use l'Hospital's rule:

$\lim f \frac{x}{g} \left(x\right) = \lim \frac{f ' \left(x\right)}{g ' \left(x\right)}$

so:

${\lim}_{x \to \infty} \ln \frac{x}{x} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} x} = {\lim}_{x \to \infty} \frac{1}{x} = 0$

Sep 25, 2017

On request, giving a demonstration without using l'Hospital:

For $x > 1$ we have:

$\ln x < x$

if we express $x$ as $x = {\left(\sqrt{x}\right)}^{2}$ then:

$\ln x = \ln {\left(\sqrt{x}\right)}^{2} = 2 \ln \sqrt{x} < 2 \sqrt{x}$

so:

$\ln \frac{x}{x} < 2 \frac{\sqrt{x}}{x} = \frac{2}{\sqrt{x}}$

On the other hand for $x > 1$ the logarithm is positive so:

$0 < \ln \frac{x}{x} < \frac{2}{\sqrt{x}}$

Then for $x \to \infty$

${\lim}_{x \to \infty} \ln \frac{x}{x} = 0$

based on the squeeze theorem.