Proving Limit Rigorously 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!
 A: 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
$$\lim_{x\to\infty}\left( \log(x)\right)^{20/x} =1 $$
A: 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 $\ln x=y\to \infty$
$$\frac{\ln (\ln x)}{\ln x}=\frac{\ln y}{y}\to 0$$
A: 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 $x\to \infty$
A: 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 $0$.
