How do I Show that, $\lim\limits_{x\to 0^+}x^{x^{x^x}} = 1$? $$\lim_{x\to 0^+}x^{(x^{(x^{x})})}$$
I tried to solve them by:
$$\lim_{x\to 0^+}\ln(x^{(x^{(x^{x})})}) = \lim_{x\to 0^+}x^{(x^{x})} \ln(x)$= \lim_{x\to 0^+}x^{(x^{x})} \cdot \lim_{x\to 0^+}\ln(x)$$
and then ends up with $0 \cdot -\infty $ (undefined)
I tried to render the graph in a web application and it shows that the result should be $1$, not undefined
What is the correct strategy to solve this limits?
 A: Hin
$ \lim_{x\to 0^+}x\ln x=0$ that is ${x\ln(x)}$ stay ner zero as $x$ is close to zero
so by Taylor expansion of first oder near zero give 
$$ x^x= e^{x\ln(x)} \sim 1+x\ln x \implies x^{x^x} = e^{x^x\ln(x)} \sim e^{(1+x\ln x)\ln(x)} = e^{\ln x+x\ln^2 x} =x e^{x\ln^2 x}$$
That is near zero we have 
 $$\bbox[yellow]{x^{x^x}  \sim  x e^{x\ln^2 x}} $$ 
But $$ x\ln^2x = [\frac{1}{2}\sqrt{x}\ln\sqrt{x}]^2\to 0 \qquad as\qquad x\to 0^+$$
Therefore, near zero we have 
 $$\bbox[yellow]{x^{x^x}  \sim  x e^{x\ln^2 x}\sim x  \qquad as \qquad x\to 0^+}$$ 
Hence, 
$$x^{x^x} \sim x\implies \lim_{x\to 0^+}x^{x^{x^{x}}} =\lim_{x\to 0^+}e^{x^{x^{x}}\ln(x)} \sim \lim_{x\to 0^+}e^{x\ln(x)} = 1$$
A: Here is a detailed expansion in $0^+$ and justification of the chain $x^{x^{x^x}}\sim x^{x^1}\sim x^x\sim 1$ 


*

*$\lim\limits_{x\to 0+}x^\alpha\ln^{\beta}x=0$ for $\alpha,\beta>0$

*$\exp(1+u)=1+u+o(u)$ when $u\to 0$



$\begin{array}{ll}
f(x) &= x^x \\
&=\exp(x\ln x)\\
&=1+O(x\ln x)&\sim 1\\
\end{array}$
$\begin{array}{ll}
g(x) &= x^{x^x}=x^{f(x)} \\
&=\exp(f(x)\ln x)\\
&=\exp(\ln x+O(x\ln ^2x))\\
&=x\exp(O(x\ln^2 x))\\
&=x(1+O(x\ln ^2x))\\
&=x+O(x^2\ln^2x)&\sim x\\
\end{array}$
$\begin{array}{ll}
h(x) &= x^{x^{x^x}}=x^{g(x)}\\
&=\exp(g(x)\ln x)\\
&=\exp(x\ln x+O(x^2\ln^3 x))\\
&=1+x\ln x+O(x^2\ln^3 x)&\sim 1\end{array}$

Note: $\frac 12x^2\ln^2 x=o(x^2\ln^3 x)$ so this term does not appear in the last line of $h(x)$ expansion.
