$0^0$ -- indeterminate, or $1$? One of my teachers argued today that 0^0 = 1. However, WolframAlpha, intuition(?) and various other sources say otherwise... 0^0 doesn't really "mean" anything..
can anyone clear this up with some rigorous explanation?
 A: Consider the following sequences:
$0,0^\frac{1}{2},0^\frac{1}{3},0^\frac{1}{4},\cdots$
$(\frac{1}{2})^0,(\frac{1}{3})^0,(\frac{1}{4})^0,\cdots$
The first evaluates to a sequence of 0s that would imply a limit of $0^0\implies0$ as each term is zero.
The second evaluates to a sequence of 1s that would imply a limit of $0^0\implies1$ as each term is one.
However, as each value is different this would imply that there isn't a value for $0^0$ as generally one would think $0^a=0$ for any positive a, while $x^0=1$ for any non-zero x as a couple of identities that conflict here.
A: Short answer: It depends on your convention and how you define exponents.
Long answer: There are a number of ways of defining exponents. Usually these definitions coincide, but this is not so for $0^0$: some definitions yield $0^0=1$ and some don't apply when both numbers are zero (leaving $0^0$ undefined).
For example, given nonnegative whole numbers $m$ and $n$, we can define $m^n$ to be the number of functions $A \to B$, where $A$ is a set of size $n$ and $B$ is a set of size $m$. This definition gives $0^0=1$ because the only set of size $0$ is the empty set $\varnothing$, and the only function $\varnothing \to \varnothing$ is the empty function.
However, an analyst might not want $0^0$ to be defined. Why? Becuase look at the limits of the following functions:
$$\lim_{x \to 0^+} 0^x = 0, \qquad \lim_{x \to 0} x^0 = 1, \qquad \lim_{x \to 0^+} (e^{-1/t^2})^{-t} = \infty$$
All three limits look like $0^0$. So when this is desired, you might want to leave $0^0$ undefined, so that it's a lack of definition rather than a discontinuity.
Typically this is resolved by:


*

*If you're in a discrete setting, e.g. considering sets, graphs, integers, and so on, then you should take $0^0=1$.

*If you're in a continuous setting, e.g. considering functions on the real line or complex plane, then you should take $0^0$ to be undefined.


Sometimes these situations overlap. For example, usually when you define functions by infinite series
$$f(x) = \sum_{n=0}^{\infty} a_nx^n$$
problems occur when you want to know the value of $f(0)$. It is normal in these cases to take $0^0=1$, so that $f(0)=a_0$; the reason being that we're considering what happens as $x \to 0$, and this corresponds with $\lim_{x \to 0} x^0 = 1$.
A: 
One of my teachers argued today that $0^0 = 1$.

He probably meant that $\displaystyle\lim_{n\to0}n^n=1$, which is indeed true.

$0^0$ doesn't really "mean" anything...

Of course it does... It's just that its meaning depends on the given context, yielding different results for different cases, as has already been explained above... As opposed to, say, $1+1$, which is always $2$, being independent on context.
