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Someone told me that $0\times \infty = 1$.

I am baffled by this because I thought you cannot multiply by infinity because it isn't a real number. If you can, is it possible to explain how and give some a proof?

Milo Brandt
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Orland
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    Zero times infinity is undefined. There's nothing to prove here. Move on. – John Dvorak Dec 27 '14 at 17:38
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    Your friend is demonstrably either wrong or using some very unconventional laws of mathematics. In particular: if $\infty + \infty = \infty$, then $$ 1 = 0 \times \infty = 0 \times (\infty + \infty) = 0 \times \infty + 0 \times \infty = 2 $$ There are other examples of why defining $0 \times \infty$ is generally troublesome, but I think this will do. – Ben Grossmann Dec 27 '14 at 17:41
  • @sas you get better results when you put the entire equality between the $ delimiters. So, in this case, it's better to write $0 \times \infty = 1$ than $0 \times \infty$ = $1$. – Ben Grossmann Dec 27 '14 at 17:44
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    An easier problem to demonstrate: we have $$ 1 = 0 \times \infty = (0 \times 0) \times \infty = 0 \times (0 \times \infty) = 0 \times 1 = 0 $$ – Ben Grossmann Dec 27 '14 at 17:47
  • @Omnomnomnom, oh... I need a smile for "facepalm" emotion. You are right, I've just missed these two dollar delimeters. Can't change this now, my fault :( – sas Dec 27 '14 at 17:47
  • @sas no big deal, just telling you in case you didn't know. – Ben Grossmann Dec 27 '14 at 17:48
  • Wrong. $0\cdot\infty=e^{-\gamma}$. – Anixx Dec 14 '21 at 05:32

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If $f(x)$ approaches $0$ and $g(x)$ approaches $\infty$ as $x$ approaches something, then $f(x)g(x)$ can approach a real number, and which number it is depends on which functions $f$ and $g$ are. For example, suppose $f(x)=(5+x)^3-5^3$ and $g(x)=\frac1x$, and consider what happens as $x\to0$. We have $$ f(x)=(5+x)^3-5^3=(125 + 75x + 15x^2 + x^3)-5^3 = 75x+15x^2+x^3. $$ Now simplify $f(x)g(x)$ and let $x\to 0$.

Michael Hardy
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  • as x tends to 0, f(x)g(x) = 75. So in some sense, 0 x infinity could equal any real number, making it undefined? – Orland Dec 27 '14 at 18:11
  • @Orland : In this context, yes. When working with integrals, one often defines $0\cdot\infty$ as $0$. For example, in $\int_{-\infty}^\infty \left.\begin{cases} e^x & \text{if }x<0 \ 0 & \text{if }x>0 \end{cases}\right},dx$, what is the integral from $0$ to $\infty$? ${}\qquad{}$ – Michael Hardy Dec 27 '14 at 18:16
  • I'm not too sure, would it equal a constant? – Orland Dec 27 '14 at 18:38
  • @Orland : It is $\int_0^\infty 0,dx=0$. Thus in this context $0\cdot\infty=0$. ${}\qquad{}$ – Michael Hardy Dec 27 '14 at 23:50
  • Is there a reason for the down-vote? – Michael Hardy Dec 28 '14 at 17:37