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

Question

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?

Answer 1

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.

Answer 2

For all $x > 0, \ln x < \sqrt x$

$\frac {\ln x}{x} < \frac {1}{\sqrt x}$

When $N > \frac {1}{\epsilon^2},$ then $x>N \implies \frac {\ln x}{x} < \frac {1}{\sqrt x} < \epsilon$