# Deconstructing $0^0$ [duplicate]

Possible Duplicate:
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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## marked as duplicate by amWhy, Arturo Magidin, Grigory M, Hans Lundmark, Zev ChonolesJun 29 '11 at 22:19

Possible duplicate: math.stackexchange.com/q/11150/9003 –  amWhy Jun 29 '11 at 21:34
why don't you check out the previous question/answers linked in my first comment, above. That addresses your question. –  amWhy Jun 29 '11 at 21:34
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 $x\to 0$. –  Arturo Magidin Jun 29 '11 at 21:37

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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mixedmath: Luke suggested an edit which I rejected. I replaced log by \log and corrected the spelling of l'Hôpital. @Luke: Could you please stop insisting on your notational fancies? Was yesterday's kerfuffle not warning enough? $\log$ is a perfectly standard notation for the natural logarithm and there is absolutely no reason to replace it by $\ln$. Learn to get a bit flexible notation-wise. –  t.b. Jun 29 '11 at 22:51
@Theo: thank you again. How do you put the carrot in l'Hôpital (without copying and pasting like I just did)? –  mixedmath Jun 30 '11 at 1:36
In answers I use the HTML-tag &ocirc; (o-circumflex), similarly &eacute; for é and &agrave; for à. In comments I just type them with my keyboard (on a mac with US keyboard I get it using alt i + o. Similarly, alt e + e for é and alt  + a for à). I think on Windows or Linux something similar using alt gr should work. (I'm used to using these things since I write often in French and German and I'm too lazy to switch my keyboard layout) –  t.b. Jun 30 '11 at 1:44
Ah, I forgot about umlauts: &auml; for ä, &ouml; for ö and &uuml; for ü. On my keyboard they're alt u + <letter>`. 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 :) –  t.b. Jun 30 '11 at 2:00

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

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@kuch nahi: I answered the OP's question "how much is $L = \lim _{x \to 0^ + } x^x$?" –  Shai Covo Jun 29 '11 at 21:50
Agreed. Ignore previous comment. –  kuch nahi Jun 29 '11 at 21:58

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 why I left the comment about this being a duplicate...it doesn't help to have even more noise about the question. –  amWhy Jun 29 '11 at 21:49
That's true, $\lim_{x\to 0^-} 0^x = \infty$. I'll correct, thank you. –  Luke Jun 29 '11 at 22:42