0
$\begingroup$

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.

  • 0
    Possible duplicate: http://math.stackexchange.com/q/11150/90032011-06-29
  • 2
    why don't you check out the previous question/answers linked in my first comment, above. That addresses your question.2011-06-29
  • 0
    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

2

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

  • 2
    that's why I left the comment about this being a duplicate...it doesn't help to have even more noise about the question.2011-06-29
  • 0
    That's true, $\lim_{x\to 0^-} 0^x = \infty$. I'll correct, thank you.2011-06-29
5

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

  • 0
    @kuch nahi: I answered the OP's question "how much is $L = \lim _{x \to 0^ + } x^x$?"2011-06-29
  • 0
    Agreed. Ignore previous comment.2011-06-29
2

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.

  • 3
    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.2011-06-29
  • 0
    @Theo: thank you again. How do you put the carrot in l'Hôpital (without copying and pasting like I just did)?2011-06-30
  • 0
    In answers I use the HTML-tag `ô` (o-circumflex), similarly `é` for é and `à` 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)2011-06-30
  • 0
    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