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Why is $\infty \cdot 0$ indeterminate?

Although $\infty$ is not a real number, it can be treated as a very large positive number, and any number multiplied by $0$ is $0$. Why not in this case?

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    $\begingroup$ Compare $\ln(x)\cdot \dfrac 1x$ to $x\cdot \dfrac 1x$ to $x^x\cdot \dfrac 1x$ as $x\to \infty$. $\endgroup$ – Bobbie D Dec 1 '16 at 4:53
  • $\begingroup$ Some related questions: (1), (2). $\endgroup$ – user 170039 Dec 1 '16 at 5:01
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    $\begingroup$ Substituting infinity for a large positive number is the same as substituting 0 for a small positive number. $\endgroup$ – Jacob Wakem Dec 1 '16 at 5:22
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    $\begingroup$ A number is "practically infinite" in a sense relative to a particular problem. There is no "practically infinite" number with respect to multiplying by 0. $\endgroup$ – Jacob Wakem Dec 1 '16 at 6:01
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Why is $\infty \cdot 0$ indeterminate?

It's called indeterminate because if you get $\infty \cdot 0$ when evaluating a limit, then you can't conclude anything about the result. Here's an example: $$ \lim_{n \to \infty} (n^2) \cdot \left(\frac{1}{n}\right). $$ Now, this limit is $\infty$. But the limit of the first thing $(n^2)$ is $\infty$, and the limit of the second $(1/n)$ is $0$. So we cannot evaluate $\infty \cdot 0$ to compute the limit.

More technically: If we have one sequence $a_1, a_2, a_3, \ldots$ of real numbers that approaches $\infty$, and another $b_1, b_2, b_3, \ldots$ that approaches $0$, the product sequence $a_1 b_1, a_2 b_2, a_3 b_3, \ldots$ might approach any real number, or it might not approach anything at all. This is what it means for $\infty \cdot 0$ to be n "indeterminate form".

Although $\infty$ is not a real number, it can be treated as a very large positive number

No, I would not agree with this statement. It may be helpful to intuitively think of $\infty$ as a very large positive number, but this is not what infinity is. $\infty$ is a sort of limit of higher and higher positive numbers. In the same way $0$ is a limit of lower and lower positive numbers.

and any number multiplied by $0$ is $0$. Why not in this case?

As explained above, $\infty$ is not a very large number, but rather a limit of larger and larger numbers. So we cannot say that multiplying $\infty$ by $0$ is $0$.

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Compare 2 sequences and their product in the next tables:

+----------+----+----+----+----+------+------------+
| sequence |    |    |    |    |      |  limit     |
+----------+----+----+----+----+------+------------+
| 1st      |  0 |  0 |  0 |  0 |  ... |  0         |
| 2nd      |  1 |  2 |  3 |  4 |  ... |  infinity  |
+----------+----+----+----+----+------+------------+
| product  |  0 |  0 |  0 |  0 |  ... |  0         |
+----------+----+----+----+----+------+------------+


+----------+----+------+------+------+------+------------+
| sequence |    |      |      |      |      |  limit     |
+----------+----+------+------+------+------+------------+
| 1st      |  1 |  1/2 |  1/3 |  1/4 |  ... |  0         |
| 2nd      |  1 |  2   |  3   |  4   |  ... |  infinity  |
+----------+----+------+------+------+------+------------+
| product  |  1 |  1   |  1   |  1   |  ... |  1         |
+----------+----+------+------+------+------+------------+


+----------+-----+------+------+------+------+------------+
| sequence |     |      |      |      |      |  limit     |
+----------+-----+------+------+------+------+------------+
| 1st      |  1  |  1/2 |  1/3 |  1/4 |  ... |  0         |
| 2nd      |  10 |  20  |  30  |  40  |  ... |  infinity  |
+----------+-----+------+------+------+------+------------+
| product  |  10 |  10  |  10  |  10  |  ... |  10        |
+----------+-----+------+------+------+------+------------+


+----------+----+------+------+------+------+-------------+
| sequence |    |      |      |      |      |  limit      |
+----------+----+------+------+------+------+-------------+
| 1st      |  1 |  1/2 |  1/3 |  1/4 |  ... |  0          |
| 2nd      |  1 |  4   |  9   |  16  |  ... |  infinity   |
+----------+----+------+------+------+------+-------------+
| product  |  1 |  2   |  3   |  4   |  ... |  infinity   |
+----------+----+------+------+------+------+-------------+

So, how to define $0 \cdot \infty$ to be consistent with all these tables?
Column limit suggests:

1st table:   $0 \cdot \infty = 0$
2nd table: $0 \cdot \infty = 1$
3rd table:  $0 \cdot \infty = 10$
4th table:  $0 \cdot \infty = \infty$

The key here is that $0$ is not necessary the limit of the sequence $0, 0, 0, \dots$, so the product $0 \cdot \infty$ is not always made by multiplying zero and a large number.

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Because $0$ does not denote the value zero, but a small number, just as $\infty$ denotes a large number. The product of a small and a large number could be anything.

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It is true that:

$$\lim_{x \to \infty} 0 \cdot x = \lim_{x \to \infty} 0 = 0$$

But this is not what you asked, does not and can not answer what you asked. Once you rationalize why it's different, you'd have answered your own question.

Hint (simplified): $0$ is a real number where usual arithmetics apply, while $\infty$ is not.

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  • $\begingroup$ maybe he actually asked so $\endgroup$ – Eric Dec 1 '16 at 12:44
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A clear answer is provided in the context of Robinson's framework where one works with hyperreal numbers. Such numbers can be infinite and infinitesimal. The poetic line that $\infty \cdot 0$ is an "indeterminate form" simply means that when one multiplies an infinite hyperreal by an infinitesimal hyperreal, the result may be either infinite, infinitesimal, or appreciable (i.e., finite but not infinitesimal). For details see Keisler's wonderful textbook Elementary Calculus which we are currently using to teach calculus to 150 freshmen.

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