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If $0^0 = 1$ what is $\lim_{x \rightarrow 0} 0^{x}?$ Intuitively, it would appear to be equal to 1 as well since $\lim_{x \rightarrow \infty} f(x) = x = 0$. At the same time, if I consider a member of the set of points approaching the left-sided limit, something like $0^{-1}$, then $0^{-1}$ would appear to be something like $\dfrac{1}{0}$ and that can't be right. S0 there would be a discontinuity at $x = -1$ for $0^{x}$. But then it appears there would be no discontinuity at $x = 0$ either then. In any case, I'm unsure how to approach answering this problem rigorously and would appreciate some advice.

Asaf Karagila
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letsmakemuffinstogether
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  • $0^0$ is not $1$, it's undefined. Your limit, though, has value $0$, since $0^x=0$ if $x\ne0$. – David Mitra Jun 07 '15 at 10:22
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    @David: There are arguments in favor of $0^0=1$, and in some contexts (although analysis is not usually one of them) it is defined as such; but all this shows is that $0^x$ is not continuous at $0$ when defining $0^0=1$. – Asaf Karagila Jun 07 '15 at 10:29
  • I don't see the point of the assumption "if $0^0=1$." Would the meaning of the question be changed if you replaced "$0^0=1$" with any random true statement, such as "$2^2=4$" or "Hobart is the capital of Tasmania"? – bof Jun 07 '15 at 10:45
  • @ bof - It's that I'm trying to provide an intuition for my hypothesis that $lim_{x \rightarrow 0} 0^{x} = 1$. – letsmakemuffinstogether Jun 07 '15 at 10:47
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    @bof: That's an odd way to extend a function to a value at $0$. "Q:What is $f(0)$? A: Hobart is the capital of Tasmania". One hell of a non-sequitur! – Asaf Karagila Jun 07 '15 at 10:48
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    OK, I guess you are not using "if" in the mathematical/logical sense, but in some rhetorical sense, as in the saying "if you're so rich why aren't you smart." In that case, you should explain why you think the fact that $0^0=1$ has something to do with your limit problem. Offhand one would seem to have nothing to do with the other: the value of $\lim_{x\to a}f(x)$ is independend of the value of $f(a)$. – bof Jun 07 '15 at 10:49
  • @AsafKaragila If you replace "$0^0=1$" with "Hobart is the capital of Tasmania" in the question title, the resulting statement is, "If Hobart is the capital of Tasmania what is $\lim_{x\to0}0^x$?". Which makes as much sense as the original, and is equivalent to it, because "truth implies P" is equivalent to P. – bof Jun 07 '15 at 10:52

2 Answers2

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Defining $0^0=1$ is a good idea when doing discrete mathematics, or set theory, or other things related to exponentiation and natural numbers.

It is less fortuitous in the context of analysis, since one can argue that $0^x$ and $x^0$ will disagree on the limit as $x\to0$, not to mention that if $x_n\to 0$ and $y_n\to 0$, then $\lim {x_n}^{y_n}$ could be a great deal of things. So it is not a bad idea to leave $0^0$ as undefined.

But looking at $\lim_{x\to0} 0^x$, we see that for every $\varepsilon>0$, there is some $\delta>0$ such that for all $x\in(0,\delta)$, $0^x=0$, and therefore this is the limit over the constant function $0$. So the limit is $0$, witnessing the discontinuity under the definition $0^0=1$.

(It should, perhaps, be remarked that $\lim_{x\to 0}x^x=1$, which a good argument in favor of this definition, but generally this exercise shows that $\lim_{(x,y)\to(0,0)}x^y$ is undefined, which supports the idea of leaving this as an undefined term.)

As a footnote, we only consider right-sided limits. Exponentiation with negative numbers involved becomes a real hassle in the domain of the real numbers. I'll leave it to you to figure out why defining this using continuity and rational sequences will not work.

Asaf Karagila
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  • Surely $0^x$ and $x^0$ agree at $x=0$. It's just that $x \mapsto 0^x$ is not continuous! Defining $x^0:=1$ for arbitrary $x$ (including $x=0$) is the only sensible choice in all areas of mathematics, not just discrete mathematics. – Martin Brandenburg Jun 07 '15 at 10:43
  • Of course, I meant the limits at $0$. And don't call me Shirley! (Thanks) – Asaf Karagila Jun 07 '15 at 10:44
  • Since in analysis it is often the case where functions are only defined on a subset of $\Bbb R$, or whatever domain you work on, I don't see why $0^0$ has to be defined. Sure, there are arguments why $0^0=1$ is the only natural choice for a definition, but this assumes that there should be a definition, and with that I don't see an issue. – Asaf Karagila Jun 07 '15 at 10:46
  • Of course we want a definition of $x^0$. For example, we want to evaluate polynomial functions $\sum_{i=0}^{n} a_i \cdot x^i$ at zero. This is $a_0$, right? This corresponds to the definiton $x^0=1$, in particular $0^0=1$. Of course we could hide this definition by writing polynomial functions as $a_0 + \sum_{i=1}^{n} a_i \cdot x^i$, but of course this is not what we want to do. – Martin Brandenburg Jun 07 '15 at 10:48
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    Martin, that depends on how you look at the polynomials. If you think about evaluating the power before evaluating the value, then there is no issue. This is a formula in the language of rings, and $x^0$ is simply a shorthand for $1$. Then there's no problem. – Asaf Karagila Jun 07 '15 at 10:51
  • Again, I'm not disputing that $0^0$ is probably the right value to assign, if you want to assign a value. I'm just saying that there's no need to insist on assigning a value. Especially if it can cause some confusion (it's easier to remember something is undefined, rather than to remember the function is not continuous, and therefore $\lim {x_n}^{y_n}\neq (\lim x_n)^{\lim y_n}$). – Asaf Karagila Jun 07 '15 at 10:54
  • I have just shown via example that it doesn't make any sense if you leave $0^0$ undefined. – Martin Brandenburg Jun 07 '15 at 10:54
  • This discussion is going nowhere. Thank you for your participation. Have a nice day, and drink plenty of water. – Asaf Karagila Jun 07 '15 at 10:56
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The function $f(x)=0^x$ is constant and zero where defined (and it is defined only for $x>0$). Any limit of the function is $0$.

ajotatxe
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