0

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.

Blue
  • 75,673
Burt
  • 1,813
  • 8
    What even is $x^{\infty}$?... – Qi Zhu Dec 30 '19 at 22:56
  • 5
    Is $1^{1/0}$ equal to $1$? Is $1^{\text{a sonnet}}$ equal to $1$? – Eric Towers Dec 30 '19 at 22:57
  • 10
    $\infty$ is not a number. – Arthur Dec 30 '19 at 23:00
  • @EricTowers what is a sonnet? Also - true, $1^\frac10$ is not one, but other than that, if we make the number we are raising one to higher and higher, it should still equal one. – Burt Dec 30 '19 at 23:01
  • 3
    @Arthur That statement means nothing and therefore says nothing. – MoonLightSyzygy Dec 30 '19 at 23:07
  • "a sonnet" is an element of the universe of things satisfying the words "any power". If you had written any element of the (second factor of the) domain of the power operation, the resolution of your question would have been evident: that resolution is Arthur's comment. – Eric Towers Dec 30 '19 at 23:12
  • Pretty sure powers are defined by algebraic equations and exclude sonnets from the domain. In fact, I'm pretty sure I saw it done. – nomen Dec 30 '19 at 23:14
  • @MoonLightSyzygy It says everything. It says that $x^\infty$, including $1^\infty$, is nonsensical as an algebraic expression. – Arthur Dec 30 '19 at 23:29
  • 2
    @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
  • 1
    @Arthur That knee jerk response that some have of saying '$\infty$ is not a number' or even 'is not a real number' when they see expressions involving $\infty$ is just naivete, and it is not spreading any valid knowledge. Instead, what should be explained is what is the precise meaning with which expressions like $1^{\infty}$ are used in a given context, for example, in the context of limits. – MoonLightSyzygy Dec 30 '19 at 23:37
  • 2
    @MoonLightSyzygy I think that to someone wrestling with the entire concept of $1^\infty$, what I said is more than valid enough for a short comment, to try to steer their thought onto more healthy tracks of mind. No need to delve into mathematical philosophy and pedantry of what "a number" means or doesn't mean. If I had written that as an answer, I would welcome the criticism because answers should be held to a higher standard. – Arthur Dec 30 '19 at 23:39
  • @Arthur There is no philosophy in explaining the exact meaning of the algebraic expression $1^{\infty}$. Just a concrete mathematical definition(s). The dismissal as 'is not a number' is the same phenomenon as '$2x=1$ has no solutions', or '$x+2=1$ has no solutions', or '$x^2=2$ has no solutions', or '$x^2+1=0$ has no solutions', given at some stages in history. – MoonLightSyzygy Dec 30 '19 at 23:42
  • See https://math.stackexchange.com/questions/2671819/what-do-indeterminate-forms-mean/2671895#2671895 – Michael Hoppe Dec 31 '19 at 11:44

1 Answers1

1

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".

Eric Towers
  • 67,037
  • 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
  • @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. – Eric Towers Dec 30 '19 at 23:43