TITLE: Why does $ \lim_{x\to 0}\ x^{x^{x}}=0?$
QUESTION [7 upvotes]: I'm having a hard time understanding why this limits equals to 0.
By simply using logarithm identity I get that this limit equals 
$$ \lim_{x\to 0} \ e^{x\ln x^{x}}=\lim_{x\to 0}\ e^{x^{2}\ln x}$$ and by L'Hopital we get that the limit of the exponent is 0 and because $f(a)=e^{a}$ is continuous I get that the final limit should be 1.
I'd love your help with understanding what I did wrong.
Thank you.

REPLY [12 votes]: The mistake you made is parenthesis in the exponent.  In any case, here is another way to solve this problem:
Since $\lim_{x\rightarrow 0^+} x^x =1$, there is an interval $(0,\delta)$, $0<\delta<1$ such that $\frac{1}{2}<x^x<\frac{3}{2}$.  Then on this interval, $$x^{\frac{3}{2}}\leq x^{x^x} \leq x^{\frac{1}{2}}$$ so we see the limit is $0$ by the Squeeze Theorem.
Hope that helps,