$\frac{1}{\infty}$ - is this equal $0$?

Question

I've seen that wolfram alpha says:

$$\frac{1}{\infty} = 0$$

Well, I'm sure that:

$$\lim_{x\to \infty}\frac{1}{x} = 0$$

But does $\frac{1}{\infty}$ only make sense when we calculate it's limit? Because for me, $1$ divided by any large amount of number will be always almost zero.

Infinity is not a number, so you cannot perform your first operation. The second expression is true. — zerosofthezeta, Nov 22 '13 at 23:48
I too have a question, but I am too lazy to write it up. What does it mean when Wolfram says 1/0 = "ComplexInfinity"? — Adam, Nov 23 '13 at 00:55
@user108425 : the answers that say "$\frac{1}{\infty}$ is wrong because $\infty$ is not a number" are oversimplifying things. Since $\frac{1}{\infty}$ is not an indeterminate form, there is a way to extend the real numbers to include two additional elements $\infty$ and $-\infty$ so that what WolframAlpha says makes sense. See Hurkyl's link. On the other hand, you should not believe everything that WolframAlpha tells you (see Adam's comment). — Stefan Smith, Nov 23 '13 at 02:20
Adam, I think WolframAlpha is referring to the compactification of $\mathbf{C}$ by adding a point at infinity. — doppz, Nov 23 '13 at 02:53

score 19 · Accepted Answer · edited Apr 13 '20 at 19:13

19

The notation $\displaystyle\frac{1}{\infty}=0$ is used as a shorthand for

"As $x$ approaches infinity, the denominator blows up without bound, and hence since the numerator is constant, the value of the function approaches zero (i.e. gets arbitrarily close to zero), and hence its limit is zero."

The notation $\dfrac 1\infty$ does NOT literally mean "divide $1$ by $\infty$".

So literally, it is nonsense; taken as shorthand for the above, you'll see that notation used pretty commonly when folks evaluate limits. It's what we call "an abuse of notation."

edited Apr 13 '20 at 19:13

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answered Nov 22 '13 at 23:42

amWhy

209,954

1

So there's no sense in saying $1^{\infty}$. Why people call it undetermined? Shouldn't this mean $\lim 1^x$? – user108425 Nov 23 '13 at 01:13
@user108425 It's a shorthand. It's clear from context that they don't mean it literally but in the context on a limit, but shorten it down for convenience. – Thomas Nov 23 '13 at 01:17
@Thomas but actually the $\lim 1^x$ is indeed $1$, but they say $1^{\infty} = $ indeterminated – user108425 Nov 23 '13 at 01:19
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@user108425: When we say that $1^\infty$ is indeterminate, the meaning is that knowing that $f(x)\to 1$ and $g(x)\to\infty$ is not enough to know what the limit of $f(x)^{g(x)}$ is. Such "indeterminate forms" are generally only interesting when both operands are something that varies. – hmakholm left over Monica Nov 23 '13 at 01:32
@amWhy: Nice answer and then some +1 – Amzoti Nov 23 '13 at 01:48
*I* mean $1/\infty$ literally. Even when talking about limits. Division is continuous there: just take the limit of a continuous function by plugging the numbers in, no need to make things all complicated. – Nov 25 '13 at 06:10

Adam · Answer 2 · 2013-12-13T21:42:53.053

Let's say that you want to divide $1$ pie among an $\infty$ of your friends. Each friend will get $\frac{1}{\infty}$ of the pie or, as you say $0$. After you give out the pie, you rethink matters and say to your friends to give the pie back to you. Each friend will give you his portion of the pie that is a $0$. So you end up with $0$ pies. Where has the pie gone?

A rather interesting argument against $0\cdot\infty =1$ occurred to me. Suppose $0\cdot\infty =1$ holds. Then $1 = \infty \cdot 0 = \infty \cdot (0 \cdot 0) = (\infty \cdot 0) \cdot 0 = 1 \cdot 0 = 0$. Strange, isn't it?

score 4 · Answer 3 · answered Nov 23 '13 at 00:10

Yes and no. Think about the implications. Is $0\cdot\infty=1$ ? Because normally, if $\displaystyle\frac ab=c$ , then $a=bc$ But this isn't really the case here, is it ? Because, since all limits of the form $\displaystyle{\lim_{n\to\infty}\frac kn}$ are $0$, for all finite numbers k, then the product $0\cdot\infty$ becomes meaningless. Sometimes it can even be infinity itself, since $\displaystyle{\lim_{n\to\infty}\frac n{n^2}=0}$, for instance. So it is best to avoid such expressions, especially if you are a beginner. Otherwise you'll soon be asking questions like why $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$ instead of $1$, since, by all appearances, $\frac1\infty=0$, and $\displaystyle\lim_{n\to\infty}1^n=1$.

$\frac{1}{\infty}$ - is this equal $0$?

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