4
$\begingroup$

Find the limit

$$\large \lim_{x\to \infty}(\ln x)^{\frac{20}x}$$

I understood that as x approached infinity, $20/x$ approached 0. This would mean that the limit would tend toward $1$. However, $\ln x$ also approaches infinity as $x$ approaches infinity. Thus, I suspected the answer to be $1$ (and it indeed is the answer), however I feel like this answer is not sufficiently rigorous. How could I rigorously prove $1$ as the answer? Any ideas/hints would be appreciated.

Note: I am a highschooler and my teachers often tell me to take these answers in faith. Thus, I may not understand any fancy notations that may usually be used in solving limit.

Thanks!

$\endgroup$
  • 2
    $\begingroup$ High-school aside, in general you should learn to use asymptotic analysis to analyze such things, because it is not only systematic but also gives you much more information than L'Hopital, and is completely rigorous. Even if the limit does not exist, asymptotic analysis will tell you what it behaves like, without guess-work involved. $\endgroup$ – user21820 Nov 26 '18 at 8:21
10
$\begingroup$

The argument you use lacks the necessary rigour. With the same argument, you would conclude that $\bigl(\mathrm e^x\bigr)^\tfrac1x\to 1$, yet $\bigl(\mathrm e^x\bigr)^\tfrac1x$ is Euler's number $\mathrm e$!

The below solutions are fine. However, as a highschooler, you might want a rigourous proof that $\frac{\ln(\ln x)}{x}$ tends to $0$ as $x$ tends to $\infty$. Here's a sketch of a simple proof: $$\frac{\ln(\ln x)}{x}=\underbrace{\frac{\ln(\ln x)}{\ln x}}_{\begin{matrix}\downarrow\\0\\\text{(setting }u=\ln x)\end{matrix}}\!\!\underbrace{\frac{\ln x}{x}}_{\begin{matrix}\downarrow\\0\end{matrix}} $$

Edit: A proof that $\lim_{x\to\infty}\dfrac{\ln x}x=0$.

For any $t>1$, $\:\sqrt t <t$, so $\dfrac 1t<\dfrac 1{\sqrt t}$, therefore $$\frac{\ln x}x=\frac1x\int_1^x \frac 1t\,\mathrm dt \le\frac1x\int_1^x \frac 1{\sqrt t}\,\mathrm dt=\frac1x(2\sqrt x-2)<2\frac1{\sqrt x},$$ and the latter tends to $0$.

$\endgroup$
  • $\begingroup$ Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method. $\endgroup$ – Dude156 Nov 26 '18 at 1:02
  • $\begingroup$ For the limit at infinity of $ \frac{\ln x}x$? Please see if the proof I've added is fine for you. $\endgroup$ – Bernard Nov 26 '18 at 1:23
9
$\begingroup$

It suffices to use the inequalities $1\le \log(x)\le x$ for $x\ge e$. Since the exponential function is increasing, we see that

$$1\le \left( \log(x) \right)^{20/x}\le x^{20/x}$$

Application of the squeeze theorem yields the coveted limit

$$\lim_{x\to\infty}\left( \log(x)\right)^{20/x} =1 $$

$\endgroup$
4
$\begingroup$

We can use that

$$\large (\ln x)^{\frac{20}x}= e^{\left[20\frac{\ln(\ln x)}{x}\right]}$$

and since by standard limits

$$\frac{\ln x}x\to 0$$

we also have

$$\frac{\ln (\ln x)}x\le \frac{\ln x}x\to 0$$

A standard way to prove the standard limits let $y=e^x\to \infty$ then

$$\frac{\ln x}x=\frac{\ln (e^y)}{e^y}=\frac y{e^y}\to 0$$

indeed eventually $e^y\ge y^2$ and then

$$\frac y{e^y}\le \frac{y}{y^2}=\frac1y\to 0$$

As an alternative we can also proceed by

$$\large (\ln x)^{\frac{20}x}=\left[(\ln x)^{1/\ln x}\right]^{\frac{20\ln x}x}\to 1$$

indeed

  • $(\ln x)^{1/\ln x}\to 1$
  • $\frac{20\ln x}x\to 0$

and the first one can be proved by

$$(\ln x)^{1/\ln x}=e^{\frac{\ln (\ln x)}{\ln x}}\to 1$$

since by $\ln x=y\to \infty$

$$\frac{\ln (\ln x)}{\ln x}=\frac{\ln y}{y}\to 0$$

$\endgroup$
3
$\begingroup$

First compute the limit of the logarithm of the expression. Indeed, $$ \frac{20}{x}\times \ln(\ln x)\to 0 $$ as $x\to \infty$. You can see this intuitively since $x$ grows much faster than $\ln(\ln x)$ or you can use L'Hospital's rule. In any case $$ \exp\left(\frac{20}{x}\times \ln(\ln x)\right)\to 1 $$ as $x\to \infty$

$\endgroup$

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.