Limit $\lim_{x\to 0} x^{x^x}$ What is:
$$\lim_{x→0} x^{x^x}$$
I'm getting 0 as an answer, but I also got infinity as an answer.
How would one solve this?
 A: If both limits exist, and the result isn't an indeterminate form, $\lim_{n\to\infty} x_n^{y_n} = \left( \lim_{n\to\infty} x_n \right)^\left(\lim_{n\to\infty} y_n\right)$.
In particular, $\lim_{x\to0^+} x^{x^x} = \left( \lim_{x\to0^+} x \right)^\left(\lim_{x\to0^+} x^x\right) = 0^1 = 0$.
A: Let $y = x^{x^x}$ then $\ln y = x^{x}\ln x $.
Since $ \lim_{x\to0^+} x^x = 1 $
We have
$$
\lim_{x\to0^+} \ln y = \lim_{x\to0^+} x^{x}\ln x = -\infty
$$
Thus
$$
\lim_{x\to0^+} y = \lim_{x\to0^+} e^{\ln y} = 0
$$
A: Using Taylor series around $x=0^+$ $$x^{x^x}=x+x^2 \log ^2(x)+\frac{1}{2} x^3 \left(\log ^4(x)+\log ^3(x)\right)+O\left(x^4\right)$$ then the limit is effectively $0$.
A: I won't be giving a rigorous answer here, I'd just be giving a rough solution:
$$\lim_{x \to 0} x^{x^x} = \, ?$$
What if we have $\lim_{x \to 0} x^x$ what do we get?
$$\lim_{x \to 0} x^x = 0^0 = 1$$
So then we'd have $\lim_{x \to 0} x^1$:
$$\lim_{x \to 0} x^1 = 0^1 = 0$$
A: Definition of a Tetration

For any positive real $x\gt 0$ and non-negative integer $n$, we have
  $$\underbrace{x^{x^{\cdot^{\cdot^x}}}}_{n}=\ ^n x=\begin{cases}
1, & n=0\\
x^{\ ^{(n-1)} x}, & n\gt 0\\
\end{cases}$$

The $n^{\rm{th}}$ tetration of $0$ is not consistently defined. However, the limit of the $n^{\rm{th}}$ tetration of $x$ as $x$ approaches zero from the right is well defined. In general we have

$$\lim\limits_{x\to 0^+} \underbrace{x^{x^{\cdot^{\cdot^x}}}}_{n}=\lim\limits_{x\to 0^+} \ ^n x=
\begin{cases}
1, & n\ \mbox{is even}\\
0, & n\ \mbox{is odd}\\
\end{cases}$$

Therefore
$$\lim\limits_{x\to 0^+} x^{x^x}=\lim\limits_{x\to 0^+} \ ^3 x=0$$
