Use $y=x^{\frac{1}{x}}$ graph and think the following calculate. $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}\ldots }}$$ I want to know how many $\sqrt{2}^{\sqrt{2}^{\sqrt{2}\ldots }}$ will be.
When $\sqrt{2}^{\sqrt{2}^{\sqrt{2} }}$ =$2$
I can't solve this calculate with using graph.
Let $a_1=\sqrt2$ and $a_n=\sqrt2^{a_{n-1}}$.
First of all, the $a_n$ is bounded above by $2$. This is clearly true when $n=1$. For $n>1$, suppose $a_{n-1}<2$. Then $a_n=\sqrt2^{a_{n-1}}<\sqrt2^2=2$. By induction, $a_n<2$ for all $n$.
Next, I claim this sequence is increasing. This is clear when $n=1$. When $n>1$, suppose $a_n>a_{n-1}$. Then $a_{n+1}=\sqrt2^{a_n}>\sqrt2^{a_{n-1}}=a_n$. By induction $a_n<a_{n+1}$ for all $n$.
Thus, $\lim_{m\to\infty}a_n$ exists. Call it $x$. Then $x=\lim_{n\to\infty}a_{n+1}=\sqrt2^{\lim_{n\to\infty}a_n}=\sqrt2^x$. Since $x\le 2$, the only solution to this is $x=2$.
P.S. The solution $x=2$ is unique since for $x\le 2$, we have $(x^{1/x})'=-x^{1/x-2}(\ln(x)-1)<0$.
P.P.S. The value of $\lim_{n\to\infty}a_n$ changes by what value you choose $a_1$ to be (with the same recursive definition). What we have seen above is that for $a_1=\sqrt2$, we have $\lim_{n\to\infty}a_n=2$. However, clearly, if $a_1=4$, $\lim_{n\to\infty}a_n=4$. In fact, it can be shown that for $a_1<4$ the limit is $2$ and for $a_1=4$ the limit is $4$. For $a_1>4$ the limit diverges.