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What is $$\lim_{x \rightarrow 0} x^0$$? Would this equal to $\lim_{x \rightarrow 0}x^x = 1$?

If the limit is undefined, would $\lim_{x \rightarrow 0^+} x^0$ be defined?

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  • $\begingroup$ what is this question? $x^0 = 1$ as x approaches 0 $x^0$ is equal to 1 $\endgroup$ Dec 25, 2014 at 5:27
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    $\begingroup$ Excellent discussion of different viewpoints here askamathematician.com/2010/12/… $\endgroup$
    – Richard
    Dec 25, 2014 at 5:32
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    $\begingroup$ @Richard I dislike this opening sentence for the mathematicians side: "Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true." There are solid reasons mathematicians choose to define it as $1$. In particular, when dealing with natural numbers, it makes absolute and simple sense to define $x^y$ inductively as $x^0=1$ and $x^{n+1}=x\cdot x^n$. To leave $0^0$ undefined leads to complexity that is pointless. Also, all empty products are $1$, again, because that is the simplest definition. An sequence of no $0$s is no different from a sequence of no 2s. $\endgroup$ Dec 25, 2014 at 5:49
  • $\begingroup$ @Richard This question is not about $0^0$. But it's natural to remark on the latter anyway, so I'll refer to Zero to the zero power - Is $0^0=1$? which offers an explanation that's both more transparent and more concise. $\endgroup$
    – user147263
    Dec 25, 2014 at 5:57
  • $\begingroup$ @ThomasAndrews I agree there are good reasons, but it has not consistently been the mathematicians choice. I recently read a paper on from less than 20 years ago that addressed the argument. The webpage gives some of the reasons for the current near consensus near the end of the article. For me, the binomial theorem was very persuasive evidence that it is the correct choice. $\endgroup$
    – Richard
    Dec 25, 2014 at 6:18

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$x^0=1$ for all $x$, in particular for $x\neq 0$. So this is the same as $\lim_{x\to 0} 1 = 1$.

$x^x$ is not well-defined on any $(-\delta,\delta)\setminus\{0\}$, so you have to be careful. $\lim_{x\to 0^+} x^x$ does exist.

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  • $\begingroup$ But isn't $0^0$ indeterminate? So it's not true that for all $x$, $x^0 =1$... $\endgroup$
    – limito
    Dec 25, 2014 at 5:31
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    $\begingroup$ But you are not considering $0^0$. The limit explains the behavior around the neighborhood of the point, but not at the point exactly. $\endgroup$
    – MathMajor
    Dec 25, 2014 at 5:33
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    $\begingroup$ Indeterminate does not mean undefined. It just means that $\lim_{(x,y)\to (0,0)} x^y$ does not exist. We use $x^0=1$ all the time in mathematics - for example, when we write $p(x)=\sum_{i=0}^n a_ix^i$ for a polynomial and say $p(0)=a_0$. $\endgroup$ Dec 25, 2014 at 5:33
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    $\begingroup$ In particular, in lambda calculus and set theory definitions of $x^y$, we get $0^0=1$ unless we go out of our way to say it is undefined. You get no contradiction by defining $0^0=1$, and we use it quite a bit, so we define it that way. We just have to remember that the function $x^y$ is not continuous there. @limito $\endgroup$ Dec 25, 2014 at 5:34
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    $\begingroup$ @Hurkyl True, but we actually use $0^0=1$ implicitly quite often. It is almost always pointless to make it undefined, as long as we recognize the discontinuity. $\endgroup$ Dec 25, 2014 at 5:52
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(i) $$\lim_{x \to 0} x^0 = 1$$

because $a^0 = 1$ for any $a \neq 0$. But with the limit, we are considering the neighborhood of $0$, and not $0$ itself.

(ii)

$$\lim_{x \to 0^+} x^x = \exp \left( \lim_{x \to 0^+} x \log x \right) \stackrel{\mathcal{L}}{=} \exp(0)=1 .$$

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  • $\begingroup$ log 0 is not defined I think. $\endgroup$
    – user4951
    May 20, 2020 at 15:23
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$$x^0=1$$was based on the assumption that $x\neq 0$ and only then we could show that $$\frac{x^m}{x^m}=x^{m-m}=x^0=1$$ Therefore since when x goes to 0 x is never 0 then $x^0$goes to 1.

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  • $\begingroup$ $x^y=\exp(y \ln x)$ for $x > 0$ and $y \in \mathbb R$ $\endgroup$
    – Zbigniew
    Dec 25, 2014 at 10:15
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Answer:

$$\lim_{x \to 0} x^0 = 1 $$

because anything to the zero power will equal 1.

/edit

$\lim_{x\to 0^+} x^x$ does exist.

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  • $\begingroup$ You mean anything $\neq 0$? $\endgroup$
    – MathMajor
    Dec 25, 2014 at 5:41
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    $\begingroup$ But $\lim_{x \to 0^+} x^x$ does exist. It equals one. Use Log-Exp technique to see this from right hand side. $\endgroup$
    – MathMajor
    Dec 25, 2014 at 5:42
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    $\begingroup$ Yes, @Bot is wrong about the limit of $x^x$. Not clear why he put quotes around "exist," there. $\endgroup$ Dec 25, 2014 at 5:42
  • $\begingroup$ I'm very skeptical about $0^0$, there was a big thread about this a few days ago with other indeterminate forms and the general consensus was that it was undefined ... ? $\endgroup$
    – MathMajor
    Dec 25, 2014 at 5:43
  • $\begingroup$ The general consensus is wrong. Mathematicians define it as $1$. Sorry if that bugs you, but that is the definition. Again, see my comments below. $\endgroup$ Dec 25, 2014 at 5:44

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