The limit of ${x^{x^x}}$ as $x\to 0^+$

Question

I wanted to determine the following: \begin{align*} \lim_{x\to 0^+}{x^{x^x}} &= 0 \end{align*} There has been a previous question posted on it before but the arguments on there do not seem entirely formal: Limit of ${x^{x^x}}$ as $x\to 0^+$

I wanted to see if we could use the fact that we know that $\lim_{x\to 0^+}{x^x} =1$

$\lim_{x\to 0^+}{x^{x^x}} = \lim_{x\to 0^+}e^{x^xlnx}=e^{{\lim_{x\to 0^+}x^xlnx}} $ but I can't see this helping me as $lnx$ has an infinite limit here.

Any thoughts as to how to get a nice solution to this?

I wouldn't say the limit of $\ln x$ as $x \to 0^+$ is indeterminate, but rather than it is $-\infty$. — Stefano, Mar 26 '17 at 11:12
@user258521 Why not prove that rigorously as a nice exercise? :-) — Simply Beautiful Art, Mar 26 '17 at 11:21

score 0 · Accepted Answer · answered Mar 26 '17 at 11:20

Note that $a* \pm \infty$ with $a\in \mathbb{R}$ is not an indeterminate form! Also notice that $\lim_{x\to 0^+} \ln(x) = -\infty$

Hence:

$$\lim_{x\to 0^+}{x^{(x^x)}} = \lim_{x\to 0^+}e^{x^xlnx}=e^{{\lim_{x\to 0^+}x^xlnx}} = e^{{\lim_{x\to 0^+}x^x * \lim_{x\to 0^+} lnx}} = e^{1*(-\infty)} = 0$$

The limit of ${x^{x^x}}$ as $x\to 0^+$

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