I am having difficulty determining is the solution for the following problem:

$$\displaystyle \lim_{x \rightarrow \infty}\left( x \times 0 \right)$$

To clarify, this question assumes ${0}$ is a constant and is absolutely zero ("true zero"), and not another figure approaching or is approximately zero ("near zero"). Thus, the question is not asking what "near zero" times "near infinity" is.

I know that ${\infty *0}$ is undefined, however my difficulty is that I'm unsure whether the answer to the problem is undefined because ${\infty *0}$ is undefined.

From my understanding, a limit does not ever 'reach' infinity - it only approaches infinity, thus there are a rational amount of numbers. As ${x\cdot 0=0}$, when x is not ${\infty}$, it seems to me that in all cases of $x$ approaching infinity the answer could also be ${0}$.

  • 1
    $\begingroup$ Would need to evaluate the quantity in the bracket first, the way this is written. Then everything is identically zero. $\endgroup$
    – AnyAD
    Oct 22, 2018 at 8:49
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    $\begingroup$ Obviously can't apply limit laws to the product. $\endgroup$
    – AnyAD
    Oct 22, 2018 at 8:50
  • $\begingroup$ @JyrkiLahtonen I think the key difference between this question and the possible duplicate, is that this question asks about the limit where 0 is a constant, while the other question asks about a term whose limit is 0. $\endgroup$
    – Brian J
    Oct 22, 2018 at 14:16
  • $\begingroup$ @BrianJ A valid point. I copy/pasted the wrong link, sorry about that. I meant to use this, but I had several candidates. Henning Makholm's answer in my opinion covers this question as well, but you can disagree. Observe that that question was closed as a dupe of the other one. $\endgroup$ Oct 22, 2018 at 14:43
  • $\begingroup$ @gimusi So it again comes down to our incompatible views of what the purpose of the site is. Yeah. Let's delete this discussion. A good idea :-) $\endgroup$ Oct 23, 2018 at 9:48

6 Answers 6


Note that for any $x$ we have $x\cdot 0=0$ and therefore

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

  • $\begingroup$ Would this change if I didn't use parenthesis? $\endgroup$
    – Roost1513
    Oct 22, 2018 at 9:05
  • $\begingroup$ @Roost1513 I've used the parenthesis just to indicate that we are taking the limit for $(x\cdot 0)$ which is the identically zero function $f(x)=0$. For that reason the limit is equal to $0$. $\endgroup$
    – user
    Oct 22, 2018 at 9:08
  • $\begingroup$ Thank you for the clarification, I just wanted to be absolutely certain. $\endgroup$
    – Roost1513
    Oct 22, 2018 at 9:09
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    $\begingroup$ @Roost1513 A similar case to be considered is $\lim_{x\to 0} \frac x x$. In that case we have an indeterminate form $0/0$ at $x=0$ but for $x\neq 0$ the function $f(x)=x/x=1$. Can you conclude what the limit is? $\endgroup$
    – user
    Oct 22, 2018 at 9:11
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    $\begingroup$ @Roost1513 For the same reason when we take the limit at $\infty$ it means that we are assuming $x$ as large as we want, say for example $x=M$, but for any $M$ we have $M\cdot 0 =0$. $\endgroup$
    – user
    Oct 22, 2018 at 9:15

As others have said, $\lim_{x\to \infty} 0 \times x = \lim_{x\to\infty} 0 = 0$. I'm going to expand a bit more on "$0 \times \infty$ is undefined".

We can't do operations with $\infty$ directly, as you know. But we can do operations with "functions with limit $\infty$", and if they behave well enough then that might give us reasonable definitions of things like "$0 \times \infty$".

However, if we replace $\infty$ by "functions with limit $\infty$" then we ought to do the same with $0$, i.e. replace $0$ by "functions with limit $0$". This is a reasonable thing to do, because it works for real numbers:

For any functions $f(x)$ and $g(x)$ such that $\lim_{x \to \infty} f(x) = a$ and $\lim_{x \to \infty} g(x) = b$, where $a$ and $b$ are real numbers (i.e. finite), it's the case that $\lim_{x \to \infty} (f(x)g(x)) = ab$.

In fact, it's better than that; if $a > 0$ is real then it's reasonable to say "$a \times \infty = \infty$":

For any functions $f(x)$ and $g(x)$ such that $\lim_{x \to \infty} f(x) = a$, where $a$ is a real number and $a > 0$, and such that $\lim_{x \to \infty} g(x) = \infty$, then $\lim_{x \to \infty}(f(x)g(x)) = \infty$ also.

However, if $f(x)$ and $g(x)$ are such that $\lim_{x \to \infty} f(x) = 0$ and $\lim _{x \to \infty} g(x) = \infty$, then we don't know anything about $\lim_{x \to \infty} f(x)g(x)$. If $g(x) = x$, then taking $f(x) = 0$, $f(x) = a/x$ (where $a > 0$), and $f(x) = 1/\sqrt x$, gives $\lim_{x \to \infty} (f(x)g(x))$ to be, respectively, $0$, $a$ and $\infty$. This is why we say "$0 \times \infty$" is undefined.

  • 3
    $\begingroup$ I especially like this answer for its explanation of what it means to say $0\times\infty$ is undefined, and why the OP or readers shouldn't take that as a fundamental rule. Though I think the formatting would be improved by making a few of the equations display-style; right now it's kind of hard to parse. $\endgroup$
    – David Z
    Oct 22, 2018 at 21:05
  • $\begingroup$ In other words, because $0\cdot\infty$ is undefined, we can't solve the problem by starting $$\lim_{x\to \infty} x \cdot 0 = (\lim_{x\to \infty} x) \cdot (\lim_{x \to \infty} 0).$$ But since we can solve it in a different way (by simplifying $x\cdot 0$ to $0$), that's okay. $\endgroup$ Oct 22, 2018 at 22:51

For every $x\in \mathbf R$ it holds $0 \cdot x = 0$. Using this definition, we let $f(x):= 0 \cdot x = 0$ for every $x\in \mathbf R$. Hence, we have $$ \lim_{x\to +\infty } x\cdot 0 = \lim_{x\to +\infty } f(x) = \lim_{x\to +\infty }0 =0$$ from the definition of the limit! There is no need to think about something like $0 \cdot (+\infty)$.


Let $f(x)=0\times x$.

Then, the following statement is true:

For every $\epsilon > 0$, there exists some $M\in \mathbb R$ such that, for all $x>M$, $|f(x) - 0| < \epsilon$.

Therefore, by the definition of a limit, we can conclude that

$$\lim_{x\to\infty} f(x) = 0$$

If you want the proof of the statement in yellow above:

Let $\epsilon > 0$ be arbitrary. Then, let $M=1$. Let $x>M$ be arbitrary. Then, $|f(x)-0| = |0\times x - 0| = |0-0|=0<\epsilon$.


When you evaluate a limit you don't actually have to worry about the value of the function at the index. In other words, when you evaluate $$\lim_{x\rightarrow x_0}f(x),$$ it is not necessary for $f(x_0)$ to be defined. A limit tells you the 'best guess' for $f(x_0)$ based on its surroundings. Take the function $$f(x)=\frac{x-3}{x-3};$$ it's not hard to see that $f$ evaluates to 1 for any $x$ except for $x=3$, where it is undefined. Based on the surroundings of $x=3$ it is possible though to give a sensible value to $f(3)$. You can find this value using a limit: $$\lim_{x\rightarrow 3}\frac{x-3}{x-3}=1$$ The limit yields 1 since you can get $f(x)$ arbitrarily close to 1 by picking $x$ arbitrarily close to $3$ (in this case it doesn't actually matter if you get closer but still).

To tie this back to your question: the limit is zero, because for every $x$ smaller than infinity you have that $x\cdot 0$ is arbitrarily close to $0$. It doesn't matter whether $\infty\cdot 0$ makes sense or not. That is why limits are so useful.


The limit is zero.

The reason is as follows. The function you are considering is $f(x) = x \times 0$. But this means that $f(x)=0$ for all real $x$. The limit of this function as x tends to infinity is 0, even though as you point out $0\times\infty$ is undefined (but we do not need to calculate that here).

Formally, to show that this limit is zero, we need to show that for all $\epsilon>0$ there exists a real $N$ so that $|f(x)-0|<\epsilon$ for all $x\ge N$. But this is trivially true for any real $N$ and any $\epsilon>0$.


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