It is well known that $0^0$ is an indeterminate form. One way to see that is noticing that
$$\lim_{x\to0^+}\;0^x = 0\quad,$$
yet,
$$\lim_{x\to0}\;x^0 = 1\quad.$$
What if we make both terms go to $0$, that is, how much is
$$L = \lim_{x\to0^+}\;x^x\quad?$$
By taking $x\in \langle 1/k\rangle_{k\in\mathbb{N*}}\,$, I concluded that it equals $\lim_{x\to\infty}\;x^{-1/x}$, but that's not helpful.
Answer
This is, unfortunately, not very exciting. Rewrite $x^x$ as $e^{x\log x}$ and take that limit. One l'Hôpital later, you get 1.
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