TITLE: Why can $2^3$ be defined but $0^0$ cannot
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: 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 $y=0$.