Deconstructing $0^0$

Question

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Zero to zero power

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 1

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.

Answer 2

$$ x^x = (e^{\ln x} )^x = e^{x\ln x} $$ tends to $1$ as $x \to 0^+$, since $x \ln x \to 0$.

Answer 3

This has been talked ad nauseam. $0^0$ is indeterminate as a limit, in the sense that $x^y$ when both $x$ and $y$ tend to zero, can tend to anything. What happens to $\lim_{x\to 0} x^x$ is another thing (it tends to 1, actually, but one cannon conclude nothing from that). And what is the value of $0^0$ is still another thing.

Anyway, the bit that follows "One way to see that is noticing that... " is also objectionable. The first limit is not entirely correct. What we know is not that $0^x=0$ for $x \ne 0$, but for $x >0$, hence that limit should be $\lim_{x\to 0^+} 0^x = 0$. But $\lim_{x\to 0^-} 0^x = \infty$