# I was told that any $x^{\infty}$ is undefined. Does this hold true even when $x=1$?

I was told that any $$x^{\infty}$$ is undefined. Does this hold true even when $$x=1$$?

If yes, why? $$1$$ to any power is always $$1$$? $$\infty$$ is not a number, but as the numbers get larger and larger, raising one to that power should still be one?

This really is all part of the question of what is $$\lim_{x\to\infty}1^x$$ and how to solve it. The limit is $$1$$, but it still leads to this question.

• What even is $x^{\infty}$?... – Qi Zhu Dec 30 '19 at 22:56
• Is $1^{1/0}$ equal to $1$? Is $1^{\text{a sonnet}}$ equal to $1$? – Eric Towers Dec 30 '19 at 22:57
• $\infty$ is not a number. – Arthur Dec 30 '19 at 23:00
• @Arthur That statement means nothing and therefore says nothing. – MoonLightSyzygy Dec 30 '19 at 23:07
• @Arthur You need to learn that 'number' is nothing in modern mathematics, simply because it is not a useful concept for anything. For the vestigial properties that the ancient concept of number had, anything could be a number in some context. You possibly are wanting to say is not a 'real number'. That is a concrete concept and it makes the claim true. However, clearly when one writes the perfectly valid expression $1^{\infty}$, one is not talking about real numbers. So, again saying that $\infty$ is not a real number says nothing about that expression. – MoonLightSyzygy Dec 30 '19 at 23:34

The literal expression "$$1^\infty$$" is undefined because infinity is not in the (relevant factor of the) domain of the power operation. Nothing in this expression is a limit or a sequence, so there is no sense in which something is getting larger and larger. This expression does not represent a process - it contains a "completed infinity".
The indeterminate form "$$1^\infty$$" appears when one is mentally approximating limits, commonly (but not exclusively) limits of the form $$\lim_{x \rightarrow \infty} f(x)^{g(x)}$$. The usual approach to such a form is to use the continuity of the exponential to write $$\lim_{x \rightarrow \infty} \mathrm{e}^{g(x) \ln(f(x))} = \mathrm{e}^{\lim_{x \rightarrow \infty} g(x) \ln(f(x))} \text{.}$$
Now if $$f$$ is approaching $$1$$ in our limit, $$\ln f$$ is approaching zero, so the limit in that exponent is of the form "$$\infty \cdot 0$$", and we look for cancellation opportunities and other familiar manipulations to resolve the relative rates of $$\ln f$$ going to zero and $$g$$ going to infinity.
What does this do to the expression you wrote? We would write $$\mathrm{e}^{\infty \cdot \ln 1} = \mathrm{e}^{\infty \cdot 0}$$. But this does not resolve the issue: "$$\infty \cdot 0$$" is also undefined because completed infinities are not in the domain of multiplication. So this expression is also "$$\mathrm{e}$$ to an undefined power".
• But, using l'hopital's rule, you would get that $e^0$ which is one. So, that concepts is true... – Burt Dec 30 '19 at 23:11
• @Burt : L'Hopital's rule applies to the indeterminate forms $\frac{0}{0}$ and $\frac{\pm \infty}{\pm \infty}$. There are no quotients in any expression you or I have written, so l'Hopital's rule applies to none of them. – Eric Towers Dec 30 '19 at 23:14
• L'Hopital can be applied to the form $1^\infty$. Here's a basic though contrived example.$$\lim_{x\to\infty}(1+1/x)^x=\exp\bigg( \lim_{x\to\infty}\frac{\ln(1+1/x)}{1/x} \bigg) \stackrel{\color{Brown}{\text{l'Hôpital}}}{=} \exp\bigg(\lim_{x\to\infty} \frac{1}{1+1/x} \bigg)=e$$ – Mason Dec 30 '19 at 23:41