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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?

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  • $\begingroup$ $\ln(x)\leq x$ for $x>0$ and show it for the function $\frac{1}{x}$. $\endgroup$
    – Yadati Kiran
    Nov 15, 2018 at 18:38
  • $\begingroup$ @Rustyn there is one for infinite limits $\endgroup$
    – gt6989b
    Nov 15, 2018 at 18:38
  • $\begingroup$ @gt6989b many thanks, I'll have a look $\endgroup$
    – Rustyn
    Nov 15, 2018 at 18:39
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    $\begingroup$ If you let $x = e^y$, then it's equivalent to show that $$ \lim_{y \to \infty} \frac{y}{e^y} = 0 $$ $\endgroup$
    – Ben Grossmann
    Nov 15, 2018 at 18:42
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    $\begingroup$ 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 $\log x<2\sqrt{x}$. $\endgroup$
    – Paramanand Singh ♦
    Nov 15, 2018 at 18:53

2 Answers 2

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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.

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  • $\begingroup$ Ah yes! I forgot that's how we defined log function. Thank you! $\endgroup$
    – Bunbury
    Nov 15, 2018 at 18:51
  • $\begingroup$ @Bunbury glad I could help, I made it a little bit easier now. $\endgroup$
    – Shashi
    Nov 15, 2018 at 18:58
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For all $x > 0, \ln x < \sqrt x$

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

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

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