Find $\lim_{x\rightarrow 0}x^{x^{x^x}}$ I already known how to prove that $\lim_{x\rightarrow 0}x^{x^x}=0$ and $\lim_{x\rightarrow 0}x^x=1$. I also tried to use L'Hôpital's rule for this question but it didn't work. How to find the limit? (The limit should be $1$ from the graph sketching.)
 A: Let us consider the more general case of a power tower of $x$ with $n$ entries. Define
$$f_0(x)=x$$
$$f_1(x)=x^x$$
$$f_2(x)=x^{x^{x}}$$
$$f_3(x)=x^{x^{x^{x}}}$$
$$\vdots$$
and so on. So your question is what is
$$\lim_{x\to 0} f_3(x)=?$$
Now, note that the limit does not make sense for real numbers if we approach $0$ from the left. As such, we will only consider right sided limits from here on out. We shall show that
$$\lim_{x\to 0^{+}}f_n(x)=\left\{
        \begin{array}{ll}
            0 & \quad n\text{ even}\\
            1 & \quad n\text{ odd}
        \end{array}
    \right.$$
For the base cases, note that it is obviously true for $n=0$ and you have already proved it for $n=1$ (in fact, you have already proved it for $n=2$). Before continuing, we will note a useful recursion for $f_n(x)$. That is
$$f_{n+2}(x)=x^{f_{n+1}(x)}=x^{x^{f_n(x)}}$$
Then to prove the inductive step, assume the proposition is true for $n-1\geq 1$. For $n$ even we have
$$\lim_{x\to 0^{+}}f_n(x)=\lim_{x\to 0^{+}}x^{f_{n-1}(x)}$$
Now, by our assumption
$$\lim_{x\to 0^{+}}f_{n-1}(x)=1$$
as $n-1$ is odd. Thus, we can use the continuity of $f_n(x)$ to conclude
$$\lim_{x\to 0^{+}}x^{f_{n-1}(x)}=0^1=0$$
Consider the case where $n$ is odd. Since $n-1\geq 1$ we are assured $n\geq 2$. Thus
$$\lim_{x\to 0^{+}}f_n(x)=\lim_{x\to 0^{+}}x^{x^{f_{n-2}(x)}}=\lim_{x\to 0^{+}}\exp\left(x^{f_{n-2}(x)}\ln(x)\right)$$
Since the exponential is continuous, we can move the limit inside to get
$$=\exp\left(\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)\right)$$
So we now ask, what is
$$\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)=?$$
By our inductive assumption, we know $f_{n-2}(x)$ is eventually bounded between $\frac{1}{2}$ and $\frac{3}{2}$. Thus
$$x^{1/2}\ln(x)\leq x^{f_{n-2}(x)}\ln(x)\leq x^{3/2}\ln(x)$$
However, it is well known that 
$$\lim_{x\to 0^{+}}x^{a}\ln(x)=0$$
for all $a>0$. By the squeeze theorem, this implies
$$\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)=0$$
We may finally conclude that
$$\lim_{x\to 0^{+}}f_n(x)=\exp\left(\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)\right)=\exp\left(\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)\right)=\exp(0)=1$$
and we are done. We conclude $f_3(x)$ goes to $1$ as $x$ goes to $0$.
A: $$x^x=\exp(\log(x)x)=1+x\log(x)+o(x\log(x))$$
$$\begin{align}
x^{x^x}&=\exp\left(\log(x)x^x\right)=\exp\left(\log(x)\big[1+x\log(x)+o(x\log(x))\big]\right)\\
&=\exp\left(\log(x)+x\log^2(x)+o\left(x\log^2(x)\right)\right) \\
&=x\exp\left(x\log^2(x)+o\left(x\log^2(x)\right)\right)=x(1+o(1))
\end{align}$$
Thus,
$$\begin{align}
x^{x^{x^x}}&=\exp\left(\log(x)x^{x^x}\right) \\
&=\exp\left(x\log(x)(1+o(1))\right)\to e^0=1
\end{align}
$$
using $\lim_{x\to 0^+}x\log(x)=0$.
Update: Since OP doesn't understand asymptotic arguments, I'm adding a solution with L'Hopital's rule. Let $L=\lim_{x\to 0^+}x^{x^{x^x}}$. Using the continuity of the logarithm:
$$
\begin{align}
\log L&=\lim_{x\to 0^+}\log(x)x^{x^{x}}=\lim_{x\to 0^+}\frac{\log(x)}{1/{x^{x^{x}}}} \\
&=\lim_{x\to 0^+}\frac{1/x}{-x^{-x^x+x-1}(x\log^2(x)+x\log(x)+1)} \\
&=\lim_{x\to 0^+}-\frac{1}{x^{-x^x+x}(x\log^2(x)+x\log(x)+1)} \\
&=\lim_{x\to 0^+}-\frac{x^{x^x}}{x^x(x\log^2(x)+x\log(x)+1)}
\end{align}
$$
and we know the limits of all expressions in the last line, so we can finish. To differentiate $1/{x^{x^{x}}}$, write it as $(x^{x^{x}})^{-1}$
