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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.

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    $\begingroup$ What even is $x^{\infty}$?... $\endgroup$ – Qi Zhu Dec 30 '19 at 22:56
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    $\begingroup$ Is $1^{1/0}$ equal to $1$? Is $1^{\text{a sonnet}}$ equal to $1$? $\endgroup$ – Eric Towers Dec 30 '19 at 22:57
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    $\begingroup$ $\infty$ is not a number. $\endgroup$ – Arthur Dec 30 '19 at 23:00
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    $\begingroup$ @Arthur That statement means nothing and therefore says nothing. $\endgroup$ – MoonLightSyzygy Dec 30 '19 at 23:07
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    $\begingroup$ @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. $\endgroup$ – MoonLightSyzygy Dec 30 '19 at 23:34
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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".

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  • $\begingroup$ But, using l'hopital's rule, you would get that $e^0$ which is one. So, that concepts is true... $\endgroup$ – Burt Dec 30 '19 at 23:11
  • $\begingroup$ @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. $\endgroup$ – Eric Towers Dec 30 '19 at 23:14
  • $\begingroup$ L'Hopital can be applied to the form $1^\infty$. Here's a basic though contrived example.$$\lim_{x\to\infty}(1+1/x)^x=\exp\bigg( \lim_{x\to\infty}\frac{\ln(1+1/x)}{1/x} \bigg) \stackrel{\color{Brown}{\text{l'Hôpital}}}{=} \exp\bigg(\lim_{x\to\infty} \frac{1}{1+1/x} \bigg)=e $$ $\endgroup$ – Mason Dec 30 '19 at 23:41
  • $\begingroup$ @Mason : I know that specific instantiations of the form of OP's problem can be coerced into a form amenable to L'Hopital's rule. Until OP actually gives a concrete starting point, any attempt to do so is speculation. $\endgroup$ – Eric Towers Dec 30 '19 at 23:43

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