Limit of ${x^{x^x}}$ as $x\to 0^+$ Can you please explain why
\begin{align*}
\lim_{x\to 0^+}{x^{x^x}} &= 0
\end{align*}
 A: Ley $y=x^{x}$. Then taking log on both sides, $\ln y=x\ln x$. Taking $x\to0^+$, we can use L'Hopital on $\displaystyle{\frac{\ln x}{\frac{1}{x}}}$, as both the numerator and denominator approach $\infty$. We find the limit to be $0$. Hence, $y\to 1$. 
Now take $z=x^{x^x}$. It is $0$ to the power something that approaches $1$. Hence, it is $0$. 
A: Since $\lim_{0^+}x\ln x=0$, we have 
$$
x^x=e^{x\ln x}=1+x\ln x+O(x^2(\ln x)^2)
$$
near $0^+$. Whence 
$$
x^{x^x}=\exp(x^x\ln x)=\exp(\ln x+x(\ln x)^2+O(x^2(\ln x)^3))
$$
$$
=x\cdot\exp(x(\ln x)^2)\cdot\exp(O(x^2(\ln x)^3)).
$$
Now since $\lim_{0^+}x(\ln x)^2=\lim_{0^+}x^2(\ln x)^3=0$, we get
$$
\lim_{0^+}x^{x^x}=0\cdot \exp(0)\cdot \exp(0)=0.
$$
Actually, the formula we have three lines above proves a little more namely:
$$
x^{x^x}\sim_{0^+} x.
$$
A: We can find $\lim_{x \to \infty} x^x$ from the definition of $e$ , but beware I am going to be very informal. We have $$\lim_{x \to 0} (1+x)^{\frac 1x}=e$$
From this it follows that $\lim_{x \to 0} 1+x=\lim_{x \to 0} e^x$ . We subtract $1$ from both sides and get $\lim_{x \to 0} x=\lim_{x \to 0} e^x-1$ , or $\lim_{x \to 0} x^x=\lim_{x \to 0} (e^x-1)^x$ . We see that the quantity in the parenthesis is approaching one, and if you have $(anything)^0$ then it's also equal to $1$ . So then the parenthesis and the exponent seem to conspire to make the limit equal to $1$  (I'm sure that if you really needed to be formal you could prove this with the squeeze theorem). Now on to your problem.  
I believe your question is $x^{(x^x)}$ . Let's substitute the limit we have just found into this limit. Since $\lim_{x \to 0} x^x=1$ , $\lim_{x \to 0} x^{x^x}=\lim_{x \to 0} x^1=0$
