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This question already has an answer here:

My question is - I know, $0\times anything=0$ and $anything \times \infty=\infty$.

So,what is $0 \times \infty$?

I suppose it's $0$ but why not $\infty$?

If I say that area of an indefinitely long line is $\infty*0=0$,where am I wrong?

I know upto limits and basic derivatives.

Thanks.

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marked as duplicate by Giuseppe Negro, David K, Mankind, anomaly, The Chaz 2.0 Oct 5 '15 at 18:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What about $\infty \times 1/\infty$? $\endgroup$ – Hubble Oct 5 '15 at 16:48
  • $\begingroup$ Is $0 \times \infty$ equal to $0$ or $\infty$? $\endgroup$ – Omnomnomnom Oct 5 '15 at 16:49
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    $\begingroup$ $0\times\infty$ is indeterminate for (more or less) the reason that you describe. This means that it could be $0$, it could be $\infty$ and it could be some other number. $\endgroup$ – Michael Burr Oct 5 '15 at 16:49
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    $\begingroup$ "I know, $0 * \text{anything} = 0$". You should think more carefully about what "anything" represents. Certainly it is nonsense to say, for example, $0 * \text{that cat over there} = 0$, because arithmetic does not teach us how to multiply a number times a cat. What you learn from arithmetic is that $0 * \text{any number} = 0$. However, $\infty$ is not a number. $\endgroup$ – Lee Mosher Oct 5 '15 at 16:53
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    $\begingroup$ It is undefined. In general, don't try to do arithmetic with infinity until you've learned about limits, and even then, be very careful. $\endgroup$ – Thomas Andrews Oct 5 '15 at 16:53
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Consider the sequences $a_n=n^2$, $b_n=\frac{1}{n}$, $c_n=\frac{\alpha}{n^2}$, and $d_n=\frac{1}{n^3}$.

Now, $$\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}n^2=\infty,$$ $$\lim_{n\rightarrow\infty}b_n=\lim_{n\rightarrow\infty}\frac{1}{n}=0,$$ $$\lim_{n\rightarrow\infty}c_n=\lim_{n\rightarrow\infty}\frac{\alpha}{n^2}=0,$$ $$\lim_{n\rightarrow\infty}d_n=\lim_{n\rightarrow\infty}\frac{1}{n^3}=0.$$

Therefore, we can think of $\lim_{n\rightarrow\infty} a_nb_n$ as $0\times\infty$ (and similarly for the others).

However, each of these products has a different answer

$$\lim_{n\rightarrow\infty}a_nb_n=\lim_{n\rightarrow\infty}n=\infty,$$ $$\lim_{n\rightarrow\infty}a_nc_n=\lim_{n\rightarrow\infty}\alpha=\alpha,$$ $$\lim_{n\rightarrow\infty}a_nc_n=\lim_{n\rightarrow\infty}\frac{1}{n}=0.$$

The problem is that the speed at which these sequences approaches infinity or $0$ changes and the different limits represent how fast the sequences are blowing up or approaching zero.

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    $\begingroup$ Is there any evidence in this question that the OP knows what a limit is? In any event, this explains why $0\times \infty$ is an indeterminate form, not why it is undefined. :) $\endgroup$ – Thomas Andrews Oct 5 '15 at 16:54
  • $\begingroup$ @ThomasAndrews No, there isn't (although there isn't information that the the OP doesn't know calculus). However, since this is a common question in calculus, I thought it would be useful to leave an answer for others with more experience. $\endgroup$ – Michael Burr Oct 5 '15 at 16:55
  • $\begingroup$ It explains why it should be undefined because if it were defined, then it would need to be all of $0$, $\alpha$, and $\infty$. Which would be a contradiction. $\endgroup$ – Michael Burr Oct 5 '15 at 17:09
  • $\begingroup$ Except that quite a few mathematicians define $0^0=1$, even though it is an indeterminate form. It being an indeterminate form is a big reason for leaving it undefined, but when there are other reasons to define the value (and $0^0$ is the only indeterminate form that ever gets defined, in reality) then it overrides the fact that the function isn't continuous. $\endgroup$ – Thomas Andrews Oct 5 '15 at 17:11
  • $\begingroup$ The choice of $0^0=1$ depends, very much, on context (as I am certain that we are both aware). One could certainly define $0\times\infty$ to have a value (and this would be consistent in certain contexts). However, without more details of the context, I would contend that this shows why a naive choice of a value for $0\times\infty$ is, at the very least, dangerous. $\endgroup$ – Michael Burr Oct 5 '15 at 17:16
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The key fact is that :

$0$ is a number

$anything$ is a number (in my mind and, I suppose, in your mind)

but $\infty$ is not a number. So we cannot define $\infty * anything$ or $\infty * 0$.

This last is only a shortcut or a mnemonic for a limit, that is a different thing than a number.

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There are many possible answers. Here is one of them: In measure and integration theory, $0\cdot\infty=0$ because it is most useful in that context. For example, the area of an infinitely long line is zero. I.e., $\text{width}\cdot\text{length}=0\cdot\infty=0$.

In other contexts, it may be better to leave the product undefined.

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$\infty$ is not a normal number and the rules of arithmetics only apply to normal numbers. The expression $0\times \infty$ is therefore not a sensible arithmetic expression that we can evaluate.

Instead, it is often used as a memonic when considering limits. If we have a sequence which grows without bounds (like $a_n=\{1,2,3,\ldots\}$) then we say $a_n\to \infty$ as $n\to \infty$ as the sequence grows without bounds. On the other hand for a sequence like $b_n = \{1 , \frac{1}{2},\frac{1}{3},\ldots\}$ we have $b_n\to 0$ as $n\to\infty$ as the terms approach $0$ as $n$ gets bigger and bigger.

Now if we have the product of two sequences $a_n\cdot b_n$ and ask what does $a_n\cdot b_n$ approach (if anything) when $n\to\infty$ then since $a_n\to\infty$ and $b_n\to 0$ we say that we have a limit on the form $0\cdot \infty$. In the case above this limit is just $1$ since $a_n\cdot b_n =\{1,1,1,\ldots\}$. Whenever you see $0\cdot \infty$ this is usually what is meant by it.

In this setting we can show that $0\cdot \infty$ is an indetermined form as it can be any number (or $\infty$) depending on the sequences we look at. We can find sequences $a_n\to\infty$ and $b_n\to 0$ such that $a_n\cdot b_n \to N$ for any real number $N$. Examples are given in the other answers.

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Here's a simple example.

The product $0 \times \infty$ is what we like to call an indeterminate form because at first glance and without limit analysis, we cannot find a solution.

  • $10^{-100} \times 10^{10}$ is going to be a very small number multiplied by a very big number. The answer is $10^{-90}$ which is a very small number (close to 0)

  • $10^{-10} \times 10^{100}$ is also a very small number multiplied by a very big number but the answer, unlike in the previous example, is $10^{90}$ which is a very big number.

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Maybe i should start of with: what does $\infty$ mean to you?
Heuristically: In real analysis we assign the "value" $\pm \infty$ to limits that do not converge to any real number and are not bounded.
(I suggest looking up the definition of convergence to infinity, as this was very heuristic and a counter-example to what i tried to describe in plain words can easily be found).

Anyways, once the definitions have been established - we can play around and come up with some useful THMs (Often referred to as limit artithmetics).

Let $\{a_n\} \xrightarrow {n\rightarrow \infty} \infty$ and $\{b_n\} \xrightarrow {n\rightarrow \infty} 0$. Then for any $A \in \mathbb R \setminus \{0\}$, $B \in \mathbb R$. Then:
$A \cdot a_n \xrightarrow {n\rightarrow \infty} \infty$
$B \cdot b_n \xrightarrow {n\rightarrow \infty} 0$

What about $ a_n \cdot b_n \xrightarrow {n\rightarrow \infty} ?$
Well, you can see (there are already plenty of examples in this page) that you can't tell what is the limit. You can't even say that it exists.

However, with the above "rules of arithemetics" (and knowing the commutative property)
$$\lim _{n\rightarrow \infty} b_n \cdot A \cdot a_n=A \cdot \lim _{n\rightarrow \infty} b_n \cdot a_n$$ And that is the most we can say. (Note that this is only meaningful if the limit exists).

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