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Lately, my friend and I were arguing about what $\infty / \infty$ equals.

My thinking was that $\infty / \infty = 1$, since no matter how high you go in the numerator, it would have to go equally as high in the denominator.

My friend pointed out that one is not the smallest it can go, and can be divided an infinite number of times. (Equaling $.\overline{0}$1)

Which is it? Or is infinity not even considered a real number and so the answer is really just undefined?

Azzie Rogers
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    How do you define $\infty$ and the fraction $\infty / \infty$? If there is no definition, of course you cannot say anything about this symbol. There is nothing to argue without a definition. – Martin Brandenburg Mar 02 '14 at 14:00
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    It's whatever you want it to be. Mathematicians choose to leave it undefined, because they don't find it useful to define it. Note, infinity is not, in and of itself, a number... – Thomas Andrews Mar 02 '14 at 14:02
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    $\dfrac \infty\infty$ is indeterminate. It tells us only that more work needs to be done to evaluate a limit. – amWhy Mar 02 '14 at 14:11
  • Thanks for all the comments and answers, that really helps. – Azzie Rogers Mar 02 '14 at 19:01
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    Nothing is indeterminate. Something is or is not equal to something, it cannot be indeterminate in any sense. $\infty$ does not denote anything, and $\infty/\infty$ much less. – Mariano Suárez-Álvarez Mar 02 '14 at 19:12

4 Answers4

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The following limits all have the indeterminate form of $\frac{\infty}{\infty}$, but they are not all $1$.

$$\lim \limits_{x \to \infty} \frac{x^2}{x}$$

$$\lim \limits_{x \to \infty} \frac{x}{x^2}$$

$$\lim \limits_{x \to \infty} \frac{x}{x}$$

However, if you are given $\frac{\infty}{\infty}$ without context, the value is indeterminate. Furthermore, note that $\infty$ is not a number, so it doesn't follow the standard rules of algebra.

We can take this one step further. $\lim \limits_{x \to \infty} \frac{x^2}{x}$ is infinite, and so is $\lim \limits_{x \to \infty} \frac{x^3}{x}$ -- their limits are the same. But doesn't that feel a bit strange? Wouldn't $x^3$ be "larger" because it's to the third power, not just the second? Well, now if we divide them, we get $\large{\lim \limits_{x \to \infty} \frac{\frac{x^3}{x}}{\frac{x^2}{x}}}$, which is $\infty$.

The conclusion overall being that, when comparing two infinite quantities, their relative growth rates -- "how fast they become infinite" -- must be considered.

MT_
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    You can in fact get any possible number as the limit. $\lim_{x\to\infty} \frac{\alpha x}{x} = \alpha$, and the limit obviously has indeterminate form $\frac{\infty}{\infty}$. – fgp Mar 02 '14 at 14:04
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    For $\alpha \gt 0$, @fgp. – Keba Mar 02 '14 at 16:09
  • @Keba are you saying that there is some $\alpha$ for which $\lim_{x\to\infty} \frac{\alpha x}{x} \neq \alpha$? – Ben Grossmann Mar 02 '14 at 19:13
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    Nope. But for $\alpha \le 0$ you don‘t get $\frac{\infty}{\infty}$. – Keba Mar 02 '14 at 19:28
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Through your thought experiments, you and your friend have discovered the reason that $\infty/\infty$ cannot be assigned a consistent meaning.

The "answer" you get, if any, will depend completely on the relative rate at which the numerator and denominator grow, as you have seen.

MPW
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In general, number systems exist because they usefully quantify things.

We invent the extended real numbers, which includes two extra numbers, $+\infty$ and $-\infty$, because they are fantastically useful for doing calculus! For example, it lets us understand

$$ \lim_{x \to +\infty} \frac{x^2 + 1}{x + 1} = +\infty $$

as saying that the function is converging to $+\infty$, as $x$ approaches $+\infty$.

It may be odd to think of this as a convergent limit, but with a little experience it will be quite natural. This lets us do other things; e.g. we can continuously extend the arctangent function by setting $\tan^{-1}(+\infty) = \pi/2$ and $\tan^{-1}(-\infty) = -\pi/2$. The arctangent is continuous at infinity, which lets us write the arctangent of a limit as the limit of the arctangent: e.g.

$$ \lim_{x \to +\infty} \tan^{-1} \left( \frac{x^2 + 1}{x + 1} \right) = \tan^{-1}(+\infty) = \frac{\pi}{2} $$

This example suggests something important: to get the most use out of our new numbers $\pm \infty$, we want the functions we use -- such as arithmetic -- to be continuous, so that we can reason as above.

In particular, we would like division to be continuous: e.g.

$$ \lim_{(x,y) \to (+\infty, +\infty)} \frac{x}{y} = \frac{+\infty}{+\infty} $$

Unfortunately, the limit on the left hand side doesn't exist, which means that we can't extend division to be continuous at $(+\infty)/(+\infty)$. Because of this, by far the most convenient thing to do is to define division so that this quotient is undefined.

If you're not familiar with multivariable calculus yet, the fact the limit above doesn't exist is a consequence of the fact that

$$ \lim_{t \to +\infty} \frac{t^2}{t} \neq \lim_{t \to +\infty} \frac{t}{t} $$

That said, if you were not considering division in general, but were instead considering the specific function defined by

$$ f(x) = \frac{x}{x} $$

Then we can continuously extend $f$ to $+\infty$ just like we did with the arctangent function, by setting $f(+\infty) = 1$. This is very analogous to the fact we like to extend $f$ so that $f(0) = 1$ as well.

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$\infty$, as you said in the last paragraph of your question, is not a number (natural or real), so the usual arithmetical operations on numbers are not defined on it.

This is why

$\dfrac{\infty}{\infty}$

is not equal to $1$.

Mauro ALLEGRANZA
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  • It is not true that «$\infty/\infty$ is not equal to $1$»: what is true is that $\infty/\infty$ does not mean anything at all (so it does not make any sense to claim it is equal to something else, let alone such a claim can be true) – Mariano Suárez-Álvarez Mar 02 '14 at 19:14
  • I wasn't going to say anything, but...it is also not true that ordinary arithmetic operations with $\infty$ are not defined. In many cases, they are easily extended to the extended reals $[-\infty,\infty]$; there are only a few cases where such operations can't be consistently defined. For example, if $r$ is real, then $r + \infty = \infty$, and $r-\infty = -\infty$. You have to be more careful with multiplication, since $0\cdot\infty$ must be excluded, and signs are important. Division is likewise well-defined with only a couple of exceptions. – MPW Mar 02 '14 at 19:56
  • @MPW - I personally stay with Thomas' answer in this post – Mauro ALLEGRANZA Mar 03 '14 at 10:14
  • @MPW - the usual way of writing $[-∞,+∞]$ does not means that $∞$ is a (natural or real) number; it is a shorthand for $x \in \mathbb R$. In the same way, $lim f(x) = ∞$, when $x \rightarrow x_0$, is a shorthand for : $\forall N, \exists \delta (|x-x_0| < \delta \rightarrow f(x) > N)$. 1/2 – Mauro ALLEGRANZA Mar 08 '14 at 09:48
  • @MPW - in $\mathbb N$ we have that : $\lnot \exists y \forall x(x < y)$, but clearly the "puported number" $+∞$ is the "greatest" of all numbers. 2/2 – Mauro ALLEGRANZA Mar 08 '14 at 09:52
  • @MauroALLEGRANZA: I'm speaking of actually adjoining the symbols $\pm\infty$ to the reals. $(-\infty,\infty)=\mathbb R$, but $[-\infty,\infty]=\mathbb R \cup {\pm\infty}\neq\mathbb R$. These symbols satisfy $-\infty< x < \infty$ for all $x\in\mathbb R$. Ordinary arithmetic on $\mathbb R$ can be extended to $[-\infty,\infty]$ in many, but not all, cases, as I have already mentioned. I think we are in agreement about most of this. – MPW Mar 08 '14 at 13:56
  • @MPW - I agree with you about the symbol. My only point is a "formal" issue: I cannot call it "number". If we call it number, it must be subject to all the axioms for numbers, like the one (see Peano axioms) saying that the function $S(x)$ (successor) is everywhere defined; so what about $S(∞)$, i.e. $∞+1$ ? – Mauro ALLEGRANZA Mar 08 '14 at 14:03
  • @MauroALLEGRANZA: We are in agreement on this :) – MPW Mar 08 '14 at 17:17