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From what I gather, we can't just define $0^0$ to be $0$ or $1$ or $69$ or whatever, because $\lim\limits_{x\mathop\to0}0^x=0$ and $\lim\limits_{x\mathop\to0}x^0=1$. So $0^0$ is called indeterminate

But why can we define (say) $2^3$? How do I know that if $\lim\limits_{x\mathop\to c}f(x)=2$ and $\lim\limits_{x\mathop\to c}g(x)=3$ then $\lim\limits_{x\mathop\to c}f(x)^{g(x)}=8$ for all functions $f$ and $g$? Is this true and if so is there a proof?

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    $\begingroup$ A very good question! (Someone will step in with a proper answer, but the short version is 'because log and exp are continuous functions, and because the limit of a product is the product of the limits', which you should be able to prove yourself if you know the definition of a limit) $\endgroup$ – Steven Stadnicki Apr 15 '14 at 3:58
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    $\begingroup$ Consider the fact that $\lim_{x \to c}{f(x)}^{g(x)}=\exp\lim_{x \to c}{{g(x)\log{f(x)}}}$. $\endgroup$ – solstafir Apr 15 '14 at 3:59
  • $\begingroup$ To answer your question, note that the right-hand side of solstafir's equation equals $\exp(\lim_{x\to c}g(x)\cdot\lim_{x\to c}(\log(f(x)))$ when both limits exist. They do when $f(x)\to2$ and $g(x)\to 3$. $\endgroup$ – Steve Kass Apr 15 '14 at 4:40
  • $\begingroup$ $0^0$ can be defined and has been by many people. $\endgroup$ – Anixx Nov 7 '15 at 19:05
  • $\begingroup$ It would be better to phrase the question with computed rather than defined. $0^0$ can very well be defined. $\endgroup$ – Yves Daoust Feb 11 '17 at 13:44
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$2^3$ is simple. We know how powers work and can multiply 2 by itself 3 times. We could do that for most numbers instead of 2 and 3, just by multiplying one by itself the other number of times. You obviously know that. $0^0$ is a little bit trickier. You may ask why, and the answer is that $0^0$ is what we like to call a degenerate case. A degenerate case is when a pattern that works over a lot of examples falls apart, or degenerates. Sort of like how we can have $1/3$ and $1/2$ and $1/1$ but $1/0$ cant exist. $0^x$ becomes $0$ for almost all numbers. $x^0$ becomes $1$ for almost all cases of x. Both of these work for all non-zero x's but at $0$ it collapses and it isnt that simple. For a better explanation and several viewpoints on why $0^0$ should actually be defined as 1 see Why is $0^0=1$? a previosuly asked question that discusses it quite well.

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  • $\begingroup$ Additionally, $0^0$ is an indeterminate form. This means that it cant really exist because of it not being defined at a point along the function its derived from. If you want to know more about determinate forms, see en.wikipedia.org/wiki/Indeterminate_form $\endgroup$ – Asimov Apr 15 '14 at 4:18
  • $\begingroup$ IMO, $0^0$ is an ambiguous expression, which can designate either a number (commonly defined to equal $1$, for good reasons), or an undeterminate form, which is a shorthand for a limit with the pattern $f(x)^g(x)$ where both functions tend to zero. As you know, the undeterminate limit can take any value, depending on the particular $f$ and $g$. $\endgroup$ – Yves Daoust Feb 11 '17 at 14:35
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You can define $x^n$ any way you want so it is certainly true that $2^3$ and $0^0$ can both be defined.

Some people claim that defining $0^0=1$ once and for all could lead to contradictions, but this claim is wrong. It is based on a general distrust of $0$, combined with faulty reasoning, such as: (a) arguments that require the impossible (requiring a discontinuous function to be continuous) or (b) arguments that use erroneous claims such as $0^x = 0$ for all $x$ (which is clearly wrong, try $x=-1$).

In math, every definition is based on convenience. The reason that the notation $x^4$ was introduced is because it is shorter than $xxxx$.

If $x^0$ wasn't defined as $1$ (for every $x$, including $x=0$) then that would be highly inconvenient. Because wherever you now see $x^n y^m$, you would have to replace that with: "if $n=0$ then $y^m$, and if $m=0$ then $x^n$, and if both are $0$ then $1$, and otherwise $x^n y^m$".

Nobody uses that kind of cumbersome descriptions, we simply write $x^n y^m$. The way the notation $x^n y^m$ is commonly used is only correct if we define $0^0$ as $1$. Fortunately, this definition is supported by many good arguments (e.g. the empty product rule, combinatorial arguments, set theory, etc., all of these produce the same conclusion) while the arguments against this definition use steps that would not be accepted in other contexts.

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  • $\begingroup$ "... could lead to contradictions ...": IMO this is one of the most interesting topic on this page. Can you substantiate the idea that $0^0:=1$ actually creates no contradiction ? $\endgroup$ – Yves Daoust Feb 11 '17 at 13:50
  • $\begingroup$ $0^0=1$ is already implicitly used by most mathematicians wherever you see $x^n$. Defining $0^0 = 1$ is simply acknowledging something that we're already doing; that won't create a contradiction unless a contradiction already exists without this definition. $\endgroup$ – Mark Feb 11 '17 at 14:21
  • $\begingroup$ Thanks but this is not the kind of answer I am expecting. Rather, what extension of a computation rule could this definition break ? [It is easy to find rules that aren't broken, I am after those that are.] $\endgroup$ – Yves Daoust Feb 11 '17 at 14:27
  • $\begingroup$ Defining $0^0=1$ breaks the "rule" that $0^x$ is always $0$. But that's OK because that "rule" is wrong anyway, e.g for $x=-1$. Defining $0^0=1$ does not break any correct rules and theorems. People use very strange logic to argue that $0^0$ should be undefined. For instance, the argument $0^x = 0$ for $x>0$. The logic behind that argument is that if a statement $P(x)$ is true for $x>0$ then it should also be true for $x=0$. This type of logic would be rejected in any area of mathematics except in the $0^0$ debate. $\endgroup$ – Mark Feb 11 '17 at 15:34
  • $\begingroup$ I am ok regarding $0^x$. Are there other counterexamples ? $\endgroup$ – Yves Daoust Feb 11 '17 at 15:43
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You certainly can define $0^0$. There is absolutely no limit on what you can define; the only thing that may happen is that your definition turns out not to be useful.

So you've made two examples of possible definitions of $0^0$. Let's first at the definition $0^0=69$. This definition would break the power laws (for example, $69 = 0^{0\cdot 2} \ne (0^0)^2 = 69^2$), without giving any benefit at all (I couldn't tell a single problem that wouldbe simplified by that definition).

On the other hand, the definition $0^0=1$ does simplify quite a few things, for example it ensures that the binomial formula $$(a+b)^n = \sum_{k=0}^n {n\choose k} a^k b^{n-k}$$ also works unchanged if $a$ or $b$ is $0$.

What however is not possible is to define $0^0$ in a way that $x^y$ is continuous for $x=0$ and $y=0$.

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