# Why is $\infty \cdot 0$ an indeterminate form, if $\infty$ can be treated as a very large positive number?

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?

• Compare $\ln(x)\cdot \dfrac 1x$ to $x\cdot \dfrac 1x$ to $x^x\cdot \dfrac 1x$ as $x\to \infty$. – Bobbie D Dec 1 '16 at 4:53
• Some related questions: (1), (2). – user 170039 Dec 1 '16 at 5:01
• Substituting infinity for a large positive number is the same as substituting 0 for a small positive number. – Jacob Wakem Dec 1 '16 at 5:22
• 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. – Jacob Wakem Dec 1 '16 at 6:01
• "Although ∞ is not a real number, it can be treated as a very large positive number" No, it can't. It just can't. – fleablood Dec 6 at 19:13

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 an "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 $$0$$.

• See an upvote, fix a typo, even after 2 years. ;) (+1) – Xander Henderson Dec 6 at 19:01

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.

Consider the following sequences: $$(x_n)_{n \in \mathbb{N}} = (\frac{1}{n})_{n \in \mathbb{N}}$$ $$(y_n)_{n \in \mathbb{N}} = (\frac{1}{n^2})_{n \in \mathbb{N}}$$ $$(z_n)_{n \in \mathbb{N}} = (n)_{n \in \mathbb{N}}.$$

We have that $$x_n \to 0$$, $$y_n \to 0$$ and $$z_n \to + \infty$$.

Note that $$x_n\cdot z_n \to 1$$ as $$x_n\cdot z_n = 1$$ for each $$n$$. Also note that $$y_n \cdot z_n \to 0$$ as $$y_n \cdot z_n = \frac{1}{n}$$ for each $$n$$.

So although you could see both limits as "$$0 \cdot (+ \infty)$$", they both have a different result. This is why it is called an 'indeterminated form': $$0 \cdot (+\infty)$$ could be anything (you can come up with your own examples to get any result you would like). You simply can not attach exactly one value to it.

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.

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.

• maybe he actually asked so – Eric Dec 1 '16 at 12:44

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.

The form "$$0\cdot \infty$$" is an indeterminate form because there are examples of functions $$f(x) \to 0$$ and $$g(x) \to +\infty$$ such that

1. $$f(x) g(x) \to 0$$,
2. $$f(x) g(x) \to L \in \mathbb{R}$$, and
3. $$f(x)g(x) \to \pm \infty$$.

Since knowing that a limit has the form $$0\cdot \infty$$ is insufficient to determine the value of that limit, such limits are said to have indeterminate form.

Typically, one of the first theorems which is taught when students start learning about limits is something like:

Theorem 1: Suppose that $$f$$ and $$g$$ are two functions, and that there are $$L,M\in\mathbb{R}$$ such that $$\lim_{x\to a} f(x) = L \qquad\text{and}\qquad \lim_{x\to a} g(x) = M.$$ Then $$\lim_{x\to a} f(x)g(x) = LM.$$

This theorem states that if two functions each have a limit at some point, then (1) the product of the two functions has a limit at that point and (2) the limit of the product is the product of the limits. Assuming that the hypotheses are met (i.e. both functions have finite limits at some point), then we can determine the limit of the product.

In cases where the hypotheses are not met (for example, if $$\lim_{x\to a} f(x) = +\infty$$), the theorem does not apply. However, we might still like to compute limits as though the theorem did apply (that is, we might like to determine a more relaxed set of hypotheses under which the statement of the theorem might still hold). For example,

Theorem 2: Suppose that $$f$$ and $$g$$ are two functions, that there is $$L>0$$ such that $$\lim_{x\to a} f(x) = L, \qquad\text{and}\qquad \lim_{x\to a} g(x) = +\infty.$$ Then $$\lim_{x\to a} f(x) g(x) = +\infty.$$

This theorem says, more or less, that $$L \cdot (+\infty) = +\infty$$ whenever $$L>0$$. Even though $$g$$ has no limit at $$a$$ (or, if you prefer, $$g$$ has an infinite limit at $$a$$), we can still determine how the limit of the product behaves at $$a$$. It is in this sense that we may treat infinity like a very large real number. It isn't actually a very large real number, but in this context, it behaves kind of like a very large real number.

Now, once we start treating infinity like a very large real number, it is very tempting to attempt to give the following theorem:

Non-Theorem: Suppose that $$f$$ and $$g$$ are two functions. Then $$\lim_{x\to a} f(x) g(x) = \left( \lim_{x\to a} f(x) \right) \left( \lim_{x\to a} g(x) \right).$$

If one attempts to use such a theorem, however, one should quickly realize that there are many examples where this simply doesn't work. Even if $$f$$ and $$g$$ behave relatively nicely near $$a$$ (e.g. one of them diverges to $$\pm \infty$$), it may not be possible to determine the limit. In any such case, the limit could be said to be indeterminate.

For example, let $$f(x) = x^{-1}$$ and let $$g(x) = x^{\alpha}$$. If $$\alpha > 0$$, then $$\lim_{x\to +\infty} g(x) = +\infty,$$ and $$\lim_{x\to +\infty} f(x) = 0$$ in any case. Therefore, with respect to the non-theorem above, we might try to conclude that $$\lim_{x\to a} f(x) g(x) "=" 0 \cdot (+\infty).$$ That is, the limit has the form $$0\cdot \infty$$. But note that there are (at least) three distinct cases:

1. If $$0 < \alpha < 1$$, then $$\lim_{x\to \infty} f(x) g(x) = \lim_{x\to \infty} x^{\alpha - 1} = 0,$$ since $$\alpha - 1 < 0$$.

2. If $$\alpha = 1$$, then $$\lim_{x\to \infty} f(x) g(x) = \lim_{x\to\infty} x^{0} = \lim_{x\to \infty} 1 = 1.$$

3. If $$\alpha > 1$$, then $$\lim_{x\to\infty} f(x) g(x) = \lim_{x\to \infty} x^{1-\alpha} = +\infty,$$ since $$\alpha - 1 > 0$$.

Hence knowing that the limit has the form $$0\cdot \infty$$ does not give us enough information to determine the value of that integral. Indeed, we can come up with examples like the above which take any possible value, including both positive and negative infinity. Since knowing the form of the limit is not enough to determine the value of the limit, we reasonably say that the limit is of indeterminate form.