Compute $\lim_{x\to 0^{+}}x^{x^{x}}=?$ Problem: find $\lim_{x\to 0^{+}}x^{x^{x}}$.
Solution:
$$\lim_{x\to 0^+}x^{x^{x}}=\lim_{x\to 0^+}e^{x^{x}\ln x}=\lim_{x\to 0^+}e^{e^{\ln x^{x}}\ln x}=\lim_{x\to 0^+}e^{e^{x\ln x}\ln x}$$
Now what should I do?
 A: First, $x\ln(x)$ goes to $0$.  
Hence, $e^{x\ln(x)}$ goes to $1$.  
It follows that $e^{x\ln(x)}\ln(x)$ goes to $-\infty$.  
Thus, $e^{e^{x\ln(x)}\ln(x)}$ goes to $0$.
A: There is a general limit theorem that if $f(x)\ge0$ for all $x$ near $c$ and if $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x)$ both exist and -- what's most important here -- are not both equal to $0$, then
$$\lim_{x\to c}f(x)^{g(x)}=(\lim_{x\to c}f(x))^{\lim_{x\to c}g(x)}$$
The theorem also applies to one-sided limits. (Note, the condition $f(x)\ge0$ is mainly to guarantee that $f(x)^{g(x)}$ is well defined.)
Now this theorem does not apply to $\lim_{x\to0^+}x^x$, but by L'Hopital, or whatever, we have $\lim_{x\to0^+}x\ln x=0$, so 
$$\lim_{x\to0^+}x^x=e^{\lim_{x\to0^+}x\ln x}=e^0=1$$
and now the general limit theorem does apply to $f(x)=x$ and $g(x)=x^x$ with $c=0$, since $\lim_{x\to0}x=0$ and $\lim_{x\to0^+}x^x=1$ are not both $0$.  We have, simply,
$$\lim_{x\to0^+}x^{x^x}=(\lim_{x\to0^+}x)^{\lim_{x\to0^+}x^x}=0^1=0$$
It's worth noting that the general theorem is again useless (as noted for $x^x$) if you ask about the next level, $x^{x^{x^x}}$. But when the "not both $0$" condition is met, it's a handy theorem indeed.
A: You're doing good; consider
$$
\lim_{x\to0^+}x^x\ln x=-\infty
$$
because $\lim_{x\to0^+}x^x=1$, so it is bounded in a right neighborhood of $0$.
Hence $\lim_{x\to0^+}e^{x^x\ln x}=0$.
In a different fashion, since $\lim_{x\to0^+}x^x=1$, we can find $0<\delta<1$ such that, for $0<x<\delta$,
$$
\frac{1}{2}\le x^x\le 2
$$
Note that, for $0<x<1$, if $a<b$ then $x^a<x^b$ (take logarithms), so, for $0<x<\delta$,
$$
x^2\le x^{x^x}\le x^{1/2}
$$
The squeeze theorem makes us end.
