Is $0^\infty$ indeterminate?

Question

Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is $0^\infty$ indeterminate? Or is it only 1 raised to the infinity that is?

Answer 1

No, it is zero.

Consider the function $f(x,y) = x^y$ and consider any sequences $\{(x_0, y_0), (x_1, y_1), \ldots\}$ with $x_i \to 0$ and $y_i \to \infty$. It is easy to see that $f(x_n,y_n)$ converges to zero: let $\epsilon > 0$. For some $N$, $|x_i| < \epsilon$ and $y_i > 1$ for all $i \geq N$, so $|f(x_i,y_i)| < \epsilon$ for all $i\geq N$.

More generally, as $x\to c$ and $y\to \infty$, $x^y$ converges to 0 for $|c|<1$, diverges to infinity for $c>1$, oscillates without converging for $c \leq -1$, and is indeterminate when $c=1$.

Answer 2

Any constant raised to the power $\infty$ or $\infty$ raised to the power of any constant is undoubtedly always indeterminate. This is because exponentiation $x^y$ is a binary operation from $\mathbb{S} \times \mathbb{S} \to \mathbb{S}$, where $\mathbb{S}$ is a set such that $x, y, x^y \in \mathbb{S}$. To the best of my knowledge, $\infty$ doesn't belong to any known set of numbers, which itself means neither $x$ nor $y$ can ever be $\infty$. That is to say, both $x^\infty$ and $\infty^y$ are indeterminate. Therefore, it follows that both $0^\infty$ and $1^\infty$ are indeterminate.

EDIT: Some people in the comments section are saying $1^\infty =1 $ and $0^\infty = 0$ by taking the help of limits. $\displaystyle \lim_{x\to\infty} 1^x$ is not the same as $1^\infty$. The former gives an approximate value of $1^x$ when $x$ is very close to $\infty$, but $x$ is still defined. Obviously, $\displaystyle \lim_{x\to\infty} 1^x = 1$ but $1^\infty \neq 1$. Both of them have different meanings. Same applies to the case of $0^\infty$ too. The question does not mention anything about limits. So the answer assumes that exact values of $1^\infty$ and $0^\infty$ are being referred to, which are absolutely indeterminate.