Why is $0^0$ also known as indeterminate? [duplicate]
Solution 1:
The underlying problem here is that there are several related but differently defined concepts that are all notated $a^b$:
Exponentiation $\mathbb R\times \mathbb N_0 \to \mathbb R$ defined as repeated multiplication, which actually works as $R\times\mathbb N_0\to R$ for any ring-with-identity $R$. According to this concept $0^0$ is unambiguously $1$.
Exponentiation $\mathbb R^\times \times \mathbb Z \to \mathbb R^\times$ defined by extending the above to negative powers as reciprocals. This works for any field (or for the units of any ring). This doesn't define $0^0$ because it only works for numbers that have multiplicative inverses.
Exponentiation $\mathbb R_{\ge 0}\times \mathbb Q_{\ge 0}\to \mathbb R_{\ge 0}$ defined by roots of integral powers (themselves defined as above). If one bothers to include $a=0$ in this definition we again get $0^0=1$.
Exponentiation $\mathbb R_+\times \mathbb Q\to \mathbb R_+$, as above plus reciprocals and $a^0=1$.
Exponentiation $\mathbb R_{\ge 0}\times \mathbb R_+ \to \mathbb R_{\ge 0}$ and/or $\mathbb R_+\times \mathbb R\to\mathbb R_+$ by limits of the two above choices. This cannot be extended to give a value for $0^0$, because it has no limit for $(a,b)\to(0,0)$.
Exponentiation $\{e\}\times \mathbb R \to \mathbb R_+$ defined as a special case by the exponential function, which is again defined for example by power series. This extends to $\{e\}\times \mathbb C \to \mathbb C^\times$ for complex numbers.
Exponentiation $\mathbb R_+\times\mathbb R\to \mathbb R_+$ defined by $a^b = e^{b\log a}$. This doesn't extend to $a=0$ at all, and only with difficulty to complex $a$; we get multiple values due to the non-injectivity of the complex exponential.
Fortunately these definitions agree for those $(a,b)$ pairs where more than one of the concepts is defined. That's why it doesn't usually create problems to use the same notations for all of them. Note that all the definitions that do give meaning to $0^0$ result in $0^0=1$. One way to express this would be that in practice $a^b$ denotes the union of all these functions.
In practice some disagreement can arise, however, when someone claims that $a^b$ is "not defined" for some particular $(a,b)$ just because one of the definitions doesn't work for that $a$ and $b$. This is especially the case for $a=b=0$.
When $0^0$ is said to be an indeterminate form, what that means is neither more nor less than the fact that the limit $$ \lim_{x\to a} f(x)^{g(x)} $$ cannot be evaluated by taking limits of $f(x)$ and $g(x)$ separately if $f(x)\to 0$ and $g(x)\to 0$.
In particular, the word "indeterminate" does not say anything about what happens if we do apply one of the exponentiation functions to $(0,0)$ and we haven't got those zeroes by "taking limits under the limit sign".
Solution 2:
In the context of limits, $0^0$ is an indeterminate form because if the "limitand" (don't know what the correct name is) evaluates to $0^0$, then the limit might or might not exist, and you need to do further investigation.
Outside of limits, it's best to define $0^0$ as $1$ because the empty product - the product of no numbers - is defined as one.
Just like the sum of no numbers is defined as $0$:
$\quad \displaystyle \sum_{\phi(i)}x_i = 0$
where $\phi$ is a statement/property that is false for all $i$.
In product notation:
$\quad \displaystyle \prod_{\phi(i)} x_i = 1$
where $\phi(i)$ is a statement/property that is false for all $i$.
Defining $0^0 = 1$ simplifies many formulas. Many textbooks and online/offline soruces prefer not to define $0^0$ but that's pusillanimous in my view. $0^0 = 1$ works, we should use it.