# Is $0^\infty$ indeterminate?

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?

## migrated from mathematica.stackexchange.comOct 10 '13 at 0:15

This question came from our site for users of Wolfram Mathematica.

• If we type those expressions into Mathematica, however, it tells us that 0^infinity is 0 and 1^infinity is indeterminate. – user3680 Oct 9 '13 at 23:42
• All the answers here assume $0^{\infty}$ is $0^{+\infty}$. But Wikipedia doesn't assume that. It claims that $0^{\infty}$ is an indeterminate form because $0^{+\infty}$ has the limiting value $0$, and $0^{-\infty}$ is equivalent to $1/0$, which, as talked about in the same place I linked, is "not commonly regarded as an indeterminate form because there is not an infinite range of values that $f/g$ could approach." – user236182 Sep 16 '17 at 23:47
• "Specifically, if $f$ approaches $1$ and $g$ approaches $0$, then $f$ and $g$ may be chosen so that $(1)$ $f/g$ approaches $+\infty$, $(2)$ $f/g$ approaches $−\infty$, or $(3)$ the limit fails to exist. In each case the absolute value $|f/g|$ approaches $+\infty$, and so the quotient $f/g$ must diverge, in the sense of the extended real numbers. (In the framework of the projectively extended real line, the limit is the unsigned infinity $\infty$ in all three cases.)" – user236182 Sep 16 '17 at 23:50
• "Similarly, any expression of the form $a/0$, with $a \neq 0$ (including $a = +\infty$ and $a = −\infty$), is not an indeterminate form since a quotient giving rise to such an expression will always diverge." – user236182 Sep 16 '17 at 23:50

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 $c=1$.

• Hm, why does wolfram seem to think this limit DNE? wolframalpha.com/input/?i=lim+x+-%3E0%2C+y-%3E+inf+of+x%5Ey – Alec Oct 10 '13 at 0:24
• wolfram alpha tells you why with the caveat: depends on the path in complex space. – Tyler Oct 10 '13 at 0:30
• user7530 guessed that Skylion intended "positive reals" for $x$ and $y$. Wolfram Alpha guessed that it means a limit for $x,y$ in the complex plane. Only Skylion can tell us what she meant. – GEdgar Oct 10 '13 at 0:32
• Because the question is ambiguous: Mathematica, as it implies, interprets it as a limit of the complex exponential. This is very different than user7530's interpretation as an operation on the extended real numbers, where the limit used to compute it is restricted to positive bases and real exponents. – Hurkyl Oct 10 '13 at 0:33
• Yes, I assumed that $x$ and $y$ are real, and $\infty$ is the positive real infinity. For complex numbers, it's also true that $x^y\to 0$ as $|x|\to 0$ and $y\to \infty$, where again here I mean the positive real infinity. – user7530 Oct 10 '13 at 0:37

Since the question came to the front page again, let's do a complex example, to show that (at least this time) Mathematica is not crazy...

For $t>0$, let $f(t) = t, g(t) = i/t$. Then $$\lim_{t\to 0+} f(t) = 0,\qquad \lim_{t\to 0+}g(t) = \infty \\ f(t)^{g(t)} = \exp\frac{i\;\log t}{t}$$ and $\mathrm{Re} \Big(f(t)^{g(t)}\Big)$ (pictured) does not converge as $t \to 0^+$.