Does $\infty^0=1$?

Question

I was wondering if $\infty^0=1$. Some people have told me that there is no answer; it is undefined. Others have told me that the answer is $1$, using the rule $a^0=1, \ a\neq 0$. If it is truly undefined, then why is it undefined? Please answer, thanks!

Answer 1

$\infty^0=\infty^{^\frac1\infty}=\sqrt[\infty]\infty$ . Now, $\sqrt[n]x$ is the quantity which, when multiplied n times with itself, gives x. So: which quantity, multiplied an infinite number of times with itself, yield infinity as a result ? Obviously, any real number $>1$ has this property, as well as infinity itself. So the result is rightly left undefined, since it is not unique. Or, better said, it is an entire interval (as opposed to a single value), namely $(1,\infty]$.

Answer 2

Since $\infty$ is not a real number, the expression $\infty^0$ is not interpreted like $3^0$, but rather as the following:

We are given two sequences of real numbers, $a_n$ and $b_n$, such that $\lim a_n=0$ and $\lim b_n=\infty$, and now we calculate the limit of the exponentiation, $\lim b_n^{a_n}$.

If $a_n=0$ for all $n$, then the limit will be $1$, because $b_n^0=1$ (recall that $b_n$ is a real number).

If $a_n$ is not identically $0$ for all sufficiently large $n$, then this will require a closer analysis of the terms $b_n^{a_n}$, and the rates in which they converge to their respective limits. Some examples:

1. $a_n=\frac1n$ and $b_n=n^n$, in this case $b_n^{a_n}=n$, so $\infty^0$ becomes $\infty$.
2. $a_n=-\frac1n$ and $b_n=n^n$, in this case $b_n^{a_n}$ converges to $0$.
3. $a_n=\frac1n$ and $b_n=2^n$, in which case $b_n^{a_n}=2$.

Answer 3

No. It depends on the rate at which something approaches $\infty$ and the rate at which the other something approaches $0$. For example,

$\lim_{n \rightarrow \infty} (e^{n})^{k/n}$ has indeterminate form $\infty^0$ However, the limit itself equals $e^k$. Therefore, by choosing your favorite constant, you just made $\infty^0$ "equal" any positive real number. This idea of rates is the only proper way to interpret $\infty^0$ since $\infty \not \in \mathbb{R}$.

Answer 4

Whether an expression like $\infty^0$ is defined depends on the context in which you expect the expression to be meaningful.

In calculus, $\infty^0$ is not defined, because there is no way to extend the power function $\mathrm{pow}:(a,b)\mapsto a^b$ to these values continuously. This interpretation is given by Asaf Karagila and mlg4080.

If you're a calculus student, and you're asking in the context of calculus, then stop reading now. Below this line is forbidden knowledge.

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In computer science, the designers of a language or other standard may choose to reserve $\infty^0$ as undefined behavior, or they may leave it as implementation-defined, or they may give it some definite value. If there is a definite value, it is typically $1$:

• In C99, POSIX, and Python, $a^0=1$ for all $a$, so in particular $\infty^0=1$.
• In .NET, $a^0=1$ for all $a$ other than NaN, so in particular $\infty^0=1$.

In higher mathematics, there are some extended versions of arithmetic that define operations which are reminiscent of $\infty^0$. If $a$ is an infinite ordinal number, cardinal number, or hyperreal number, then $a^0=1$. This is sas' answer. But it is uncommon to write "$\infty$" in place of $a$, since there are many different infinite numbers in each arithmetic.