# Show that $\lim\limits_{x \to \infty}\frac{\ln x}{x}=0$ from definition.

Show that $$\displaystyle\lim_{x \to \infty}\frac{\ln x}{x}=0$$.

I know this is a simple application of the L'Hopital's rule, but can we also show this from the $$\displaystyle\epsilon-\delta$$ definition?

I am stuck because while it is easy to find a lower bound for the denominator $$x$$, the numerator does not have an upper bound - it merely increases less fast than the denominator. Is there a way to manipulate the expression to get a bound?

• $\ln(x)\leq x$ for $x>0$ and show it for the function $\frac{1}{x}$. Nov 15, 2018 at 18:38
• @Rustyn there is one for infinite limits Nov 15, 2018 at 18:38
• @gt6989b many thanks, I'll have a look Nov 15, 2018 at 18:39
• If you let $x = e^y$, then it's equivalent to show that $$\lim_{y \to \infty} \frac{y}{e^y} = 0$$ Nov 15, 2018 at 18:42
• Well its just an immediate but non-obvious consequence of the inequality $\log x\leq x-1$ or if you prefer the simpler inequality $\log x<x$. The non-obvious part is related to the fact that we need to replace $x$ with $\sqrt{x}$ in above inequality to get $\log x<2\sqrt{x}$. Nov 15, 2018 at 18:53

Assuming you define the $$\ln(\cdot)$$ by the integral, then it is more or less enough to notice that $$0\leq \int^x_1\frac 1 t\, dt\leq \int^x_1\frac 1{ \sqrt t} \, dt\leq \int^x_0\frac{1}{\sqrt t} \, dt$$ This leads to $$\left |\frac{\ln(x)} {x} \right|\leq \left|\frac{2\sqrt{x}}{x} \right|$$ The rest is straightforward.
For all $$x > 0, \ln x < \sqrt x$$
$$\frac {\ln x}{x} < \frac {1}{\sqrt x}$$
When $$N > \frac {1}{\epsilon^2},$$ then $$x>N \implies \frac {\ln x}{x} < \frac {1}{\sqrt x} < \epsilon$$