1
$\begingroup$

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

How can I know this without looking at the graph? What's the easiest way to equate that expression to 0?

I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Do you know l'Hospital? $\endgroup$ – user302982 Jul 31 '17 at 7:56
  • $\begingroup$ Thanks for pointing me in some direction. I'll look into it. $\endgroup$ – StopReadingThisUsername Jul 31 '17 at 7:57
  • $\begingroup$ $\ln x=2\ln\sqrt{x}\le2(\sqrt{x}-1)$ for $x>0.$ $\endgroup$ – Professor Vector Jul 31 '17 at 7:57
  • $\begingroup$ How do we know that $ln \sqrt{x} <= \sqrt{x} - 1$? $\endgroup$ – StopReadingThisUsername Jul 31 '17 at 7:59
  • 1
    $\begingroup$ Another way to think of the result is zero is that the increment of $\ln$ is very slow. Example, $x=1000$ $\ln x=6.907755279$ So, the denominator increases faster than the numerator. By it, we can conclude it tends to zeroes as $x$ tends to $\infty$ $\frac{\ln x}{x}=\frac{6.907755279}{1000}$. Logarithmic is a very slow increasing function. Its inverse is the exponential function which has a quite fast increasing property. $\endgroup$ – Crazy Jul 31 '17 at 8:13
5
$\begingroup$

$$\lim_{x\to \infty} {\ln x \over x} =\lim_{x\to \infty} \ln x^{1/x} = \ln\left( \lim_{x\to \infty} x^{1/x}\right) = \ln 1 = 0$$

Limit used : How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?

|cite|improve this answer
$\endgroup$
  • 2
    $\begingroup$ The limit from the link is not sufficient, since it is a limit of a sequence while what you need is a limit of a function as $x \to \infty$. $\endgroup$ – Adayah Jul 31 '17 at 8:32
  • 1
    $\begingroup$ Brilliant solution! Thanks a bunch! :D $\endgroup$ – StopReadingThisUsername Jul 31 '17 at 9:22
  • $\begingroup$ @Adayah I thought that AM-GM with squeeze theorem proof will work. $\endgroup$ – user8277998 Jul 31 '17 at 19:00
  • $\begingroup$ I suggest putting it in the answer since it is not evident how to do it, at least for me. Unless you take for granted that $x^{1/x}$ is decreasing from some point, then it's clear. $\endgroup$ – Adayah Jul 31 '17 at 19:19
7
$\begingroup$

By L'Hospital rule

$$\lim_{x \rightarrow \infty}\frac{\ln x}{x}=\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{1}$$

$$\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{1}=\lim_{x \rightarrow \infty}\frac{1}{x}$$

$$\lim_{x \rightarrow \infty}\frac{1}{x}=0$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ A quick proof in just 3 lines. Nice! (+1) $\endgroup$ – Adayah Jul 31 '17 at 8:33
  • 1
    $\begingroup$ Makes sense. Thanks! $\endgroup$ – StopReadingThisUsername Jul 31 '17 at 9:23
7
$\begingroup$

Since $\ln t \le t$, for $t = \sqrt{x}$ we would have

$$\ln \sqrt{x}=\dfrac{1}{2}\ln x \leq \sqrt{x} \iff \ln x \leq 2\sqrt{x}$$

and, for $x>1$,

$$0\le\ln x\leq 2\sqrt{x}\iff 0\le\dfrac{\ln x}{x}\leq \dfrac{2}{\sqrt{x}}$$

With the squeeze theorem we have $\dfrac{\ln x}{x} \underset{x\to +\infty}{\longrightarrow}0$

|cite|improve this answer
$\endgroup$
4
$\begingroup$

With $x=e^t$ we have to investigate $\lim_{t\to \infty} \frac{t}{e^t} $. For $t>0$ we have

$0<\frac{t}{e^t}=\frac{t}{1+t+\frac{t^2}{2}+...} \le \frac{t}{\frac {t^2}{2}}=\frac{2}{t}$.

Your turn !

|cite|improve this answer
$\endgroup$
4
$\begingroup$

A high school proof, using only basic properties of the integral:

For $t\ge 1$, we have $\sqrt t\le t$, so $\dfrac1t\le\dfrac 1{\sqrt t}$ and hence, if $x\ge 1$, $$ \ln x=\int_1^x\frac{\mathrm d\mkern1mut}{t}\le\int_1^x\frac{\mathrm d\mkern1mut}{\sqrt t}=2(\sqrt x-1)<2\sqrt x$$ from which we deduce at once $$0\le\frac{\ln x}{x}<\frac{2\sqrt x}{x},\quad\text{which tends to }0.$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ one typo:$t\ge\sqrt{t}$ $\endgroup$ – Stu Jul 31 '17 at 9:03
  • $\begingroup$ Oops! Thanks for pointing it! $\endgroup$ – Bernard Jul 31 '17 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.