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

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?

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

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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By L'Hospital rule

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

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

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

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    $\begingroup$ A quick proof in just 3 lines. Nice! (+1) $\endgroup$ – Adayah Jul 31 '17 at 8:33
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    $\begingroup$ Makes sense. Thanks! $\endgroup$ – StopReadingThisUsername Jul 31 '17 at 9:23
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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 $\dfrac{\ln x}{x} \underset{x\to +\infty}{\longrightarrow}0$

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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 !

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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 $$0\le\frac{\ln x}{x}<\frac{2\sqrt x}{x},\quad\text{which tends to }0.$$

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  • $\begingroup$ 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

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