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Find the limit

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

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!

The argument you use lacks the necessary rigour. With the same argument, you would conclude that $bigl(mathrm e^xbigr)^tfrac1xto 1$, yet $bigl(mathrm e^xbigr)^tfrac1x$ 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:
$$frac{ln(ln x)}{x}=underbrace{frac{ln(ln x)}{ln x}}_{begin{matrix}downarrow\0\text{(setting }u=ln x)end{matrix}}!!underbrace{frac{ln x}{x}}_{begin{matrix}downarrow\0end{matrix}} $$

Edit: A proof that $lim_{xtoinfty}dfrac{ln x}x=0$.

For any $t>1$, $:sqrt t <t$, so $dfrac 1t<dfrac 1{sqrt t}$, therefore
$$frac{ln x}x=frac1xint_1^x frac 1t,mathrm dt lefrac1xint_1^x frac 1{sqrt t},mathrm dt=frac1x(2sqrt x-2)<2frac1{sqrt x},$$
and the latter tends to $0$.

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

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

Application of the squeeze theorem yields the coveted limit

$$lim_{xtoinfty}left( log(x)right)^{20/x} =1 $$

We can use that

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

and since by standard limits

$$frac{ln x}xto 0$$

we also have

$$frac{ln (ln x)}xle frac{ln x}xto 0$$

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

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

indeed eventually $e^yge y^2$ and then

$$frac y{e^y}le frac{y}{y^2}=frac1yto 0$$

As an alternative we can also proceed by

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

indeed

• $(ln x)^{1/ln x}to 1$


• $frac{20ln x}xto 0$

and the first one can be proved by

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

since by $ln x=yto infty$

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

First compute the limit of the logarithm of the expression. Indeed,
$$
frac{20}{x}times ln(ln x)to 0
$$
as $xto infty$. 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
$$
expleft(frac{20}{x}times ln(ln x)right)to 1
$$
as $xto infty$