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Suppose $x$ is a positive real number and $a$ an integer. On the one hand, one has $$ x^{0} = x^{a-a} = x^{a} \cdot \frac{1}{x^{a}} = 1\tag{1} $$

On the other hand, $x^{\frac{1}{3}}$ is a number such that when multiplied by itself $3$ times it gives $x$. For instance $$ (2)^{1/3}\cdot(2)^{1/3}\cdot (2)^{1/3}=2. $$

More generally, $x^{\frac{1}{m}}$ ($m$ being some positive integer) is a number such that when multiplied by itself $m$ times, it gives $x$: $$ \underbrace{x^{\frac1m}\cdot x^{\frac1m}\cdot\cdots \cdot x^{\frac1m}}_{m\text{ times}}=x\tag{2} $$

In the same way, since $\frac{1}{\infty}=0$ and thus by (1), $$ x^{\frac{1}{\infty}} =x^{0} = 1, $$ I should expect that $1$ is a number such that when multiplied by itself infinitely many times, it gives $x$, which is not true when $x\neq 1$.

Question: Something must be wrong. But how should I understand the discrepancy above?


Background assumption: the question above was asked by a high school student without having taken a rigorous course in real analysis. I'm looking for answers to students with such background.


[Added later]

The word "discrepancy" in the question refers to the following two seemingly contradictory observations by the student:

  • Using the logic in (2), one may expect that $$ \underbrace{x^{\frac1\infty}\cdot x^{\frac1\infty}\cdot\cdots \cdot x^{\frac1\infty}}_{\infty\text{ times}}=x\tag{3} $$

  • On the other hand, $\frac{1}{\infty}=0$ implies that $$ \underbrace{x^{\frac1\infty}\cdot x^{\frac1\infty}\cdot\cdots \cdot x^{\frac1\infty}}_{\infty\text{ times}}= \underbrace{1\cdot 1\cdot\cdots \cdot 1}_{\infty\text{ times}}=1\tag{4} $$

When $x\neq 1$, (3) clearly contradicts (4).

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    $\begingroup$ What I can say is that there's no $\infty$ if we don't speak language of limit. Even if there was, $1/m \neq 1/\infty$ for any $m$. They are different by a little and many little add up to the different of $x$ and $1$. $\endgroup$ – Quang Hoang Feb 6 '18 at 4:11
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    $\begingroup$ In the same way There is a "leap of faith" around this point, which is a bit hard to point out without using any concepts of real analysis at all. Something must be wrong Maybe ask the student to retrace the same steps with $\,x=0\,$, which would "prove" in the end that $\,0^{1 / \infty} = 1\,$. The latter may be easier to argue as wrong, both formally and intuitively. $\endgroup$ – dxiv Feb 6 '18 at 4:16
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    $\begingroup$ Since the previous post was handled and subsequently deleted by mods. Instead of re-posting it, it seems more reasonable to bring it on meta first and see what the community thinks about it. $\endgroup$ – user99914 Feb 6 '18 at 17:45
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    $\begingroup$ A memorandum for the future: please avoid re-posting the same question just to circumvent downvotes; please stick to the rules and try to improve the original question in order to put it in the reopening queue, if needed. $\endgroup$ – Jack D'Aurizio Feb 6 '18 at 21:50
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    $\begingroup$ (cont.) But MSE "is for people studying mathematics at any level and professionals in related fields." Now that the old bad version has gone, I don't see why such question can not exist in MSE. $\endgroup$ – Jack Feb 7 '18 at 0:49
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You can't ever really divide by $\infty$, only an arbitrarily large number $N$. (in the same way that you can't divide by $0$ but just a small $\epsilon$)

You can explain this rigorously using limits but it's probably better to tell a high school student to not divide by $\infty$ or $0$.

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  • $\begingroup$ Part of the confusion might be that one could take the limit of $\lim_{x \to \infty}{\frac{1}{x}}$ which would be equal to $0$ (but that doesn't mean that one could directly divide $\frac{1}{\infty}$. $\endgroup$ – Gil Keidar Feb 9 '18 at 17:14
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Formally, we only define finite length expressions, so multiplying a number by itself an infinite number of times does not make sense. If you talk about multiplying an infinite set of numbers you are talking informally and glossing over the taking of a limit. It is true that $\lim_{m \to \infty}x^{\frac 1m}=1$ but that does not that you can multiply an infinite number of $1$s and get $x$. You clearly cannot, as you can't tell which $x$ should be $1^{\infty}$. $\frac 1{\infty}=0$ should be viewed as an informal statement that is a useful way to think of $\lim_{z \to \infty}\frac 1z=0$ and as long as you don't get an indeterminate form you can use it to evaluate many limits.

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What you have is $(x^{\frac 1m})^m = 1$ and, as $m \to \infty$ you get, at least symbolically, $(x^0)^\infty = 1$. This is an indeterminate form. That is, depending on how you approach $0$ and how you approach $\infty$, $(x^0)^\infty$ can approach any positive number. For example $\displaystyle \lim_{n \to \infty} \left( N^\frac 1n \right)^n = N$ and $\displaystyle \lim_{n \to \infty} \left( N^\frac 1n \right)^{2n} = N^2$.

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