Wrong reasoning with $x^{\frac{1}{\infty}}=x^0=1$?

Question

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?

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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.

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[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).

Answer 1

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 $0$.

Answer 2

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.

Answer 3

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 $\displaystyle \lim_{n \to \infty} \left( N^\frac 1n \right)^{2n} = N^2$.