Though $0^0$ is an indeterminate form of limits, is it undefined?

Question

I’ve done a my research, though I have not been able to find an adequate explanation as to whether or not $$0^0$$ exists as a real number, and why or why not? I must credit this question to “Question on the controversial ‘undefined’ $0^0$.”

This Wikipedia entry lists a (seemingly?) exhaustive list of indeterminate forms of limits, with which I take no issue. All of which involve $\infty$ or a ‘multiple’ of $1/0$ and are therefore undefined as real numbers—that is, all except $0^0$.

Desmos clearly claims that $0^0=1$; however, my TI-84 returns ERROR: DOMAIN in both real and complex mode. So, which is it? Is $0^0$ defined or not?

_{I wasn’t quite sure what other tags apply, so feel free to edit.}

Answer 1

Zero to zeroth power is often said to be "an indeterminate form", because it could have several different values.

Since $x^0$ is $1$ for all numbers $x$ other than $0$, it would be logical to define that $0^0 = 1$.

But we could also think of $0^0$ having the value $0$, because zero to any power (other than the zero power) is zero.

Also, the logarithm of $0^0$ would be $0$ · infinity, which is in itself an indeterminate form. So laws of logarithms wouldn't work with it.

So because of these problems, zero to zeroth power is usually said to be indeterminate.

However, if zero to zeroth power needs to be defined to have some value, $1$ is the most logical definition for its value. This can be "handy" if you need some result to work in all cases (such as the binomial theorem).

Answer 2

Why not both? Indeterminate (in the sense of calculus) but defined (in the sense of natural numbers or cardinal numbers or ordinal numbers)

indeterminate (in the sense of calculus)
If $0 \le z \le +\infty$, then there exist sequences $a_n, b_n \in (0,+\infty)$ with $a_n \to 0, b_n \to 0$ such that $a_n^{b_n} \to z$, and therefore there also exist such sequences such that $a_n^{b_n}$ does not converge.

defined for cardinal numbers
The cardinal number $0$ represents the number of elements in the empty set. And thus $0^0=1$ means there is a unique function from $\varnothing$ to $\varnothing$. Similarly for natural numbers and ordinal numbers: the usual definition of exponentiation yields $0^0=1$.

In fact we also want to define $0^0=1$ when the exponent is the natural number $0$ and the base is the real or complex number $0$. This will let the binomial theorem, and usual notation for polynomials and power series work as expected.

undefined for floating point $0$ in the exponent
Related to the "indeterminate" case. Definition of exponentiation for complex numbers: $a^b = \exp(b\log a)$. This is undefined for $a=b=0$.

Perhaps the confusing thing is using the same symbol $0$ for all those different things!

Answer 3

The problem isn't about assigning a value to the symbol "$0^0$"; what indeterminate means in this context is that knowing that $\lim_{x\to x_0} f(x), g(x) = 0$ does not determine $\lim_{x\to x_0} f(x)^{g(x)}$. For example, taking $f\equiv 0, g(x) = x$ gives $\lim_{x\to 0} f(x)^{g(x)} = 0$ but $\lim_{x\to 0} g(x)^{f(x)} = 1$.