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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.

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    There is an essential discontinuity at $(x,y)=(0,0)$ for the function $(x,y)\mapsto x^y$; approaching along the line $x=y$ gives a limit of $1$, though, since $x\ln x\to 0$ as $x\to 0$.2011-06-29

3 Answers 3

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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 $\lim_{x\to 0^-} 0^x = \infty$

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    That's true, $\lim_{x\to 0^-} 0^x = \infty$. I'll correct, thank you.2011-06-29
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$ x^x = (e^{\ln x} )^x = e^{x\ln x} $ tends to $1$ as $x \to 0^+$, since $x \ln x \to 0$.

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    Agreed. Ignore previous comment.2011-06-29
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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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    Ah, I forgot about umlauts: `ä` for ä, `ö` for ö and `ü` for ü. On my keyboard they're `alt u + `. I often wonder why this accent-business is so difficult, especially for mathematicians. The equation $\ddot{u} = u$ is interesting, while $u = u$ is boring :)2011-06-30