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In $[0,+\infty)$, $0^+\times +\infty$ can be any number in $(0,+\infty)$ so is undetermined; (in which $0^+$ means when a variable approaches to $0$). Because $\lim_{x\rightarrow 0}\dfrac{r}{x}=+\infty$, for $r\in (0,+\infty)$. And, $0\times \infty=0$ (in which $0$ means absolute zer0).

In measure theory, we extend non-negative real axis to $[0,+\infty]=[0,+\infty)\cup{\{r_0=+\infty}\}$ (I used of a definition came in Terence Tao's book in here, page $xi$ - "... with an additional element adjoined to it, which we label $+∞$"). So what is the value of $0\times +\infty$ (in which $+\infty = r_0$ and $0$ is the absolute value)?

Detailed answers with supportive proofs would be really appreciated.

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  • $\begingroup$ Doesn't it declare $+\infty \cdot 0 = 0 = 0 \cdot +\infty$ on that same page? $\endgroup$ – pjs36 Aug 1 '15 at 3:29
  • $\begingroup$ @pjs36 - yes, but didn't make sense for me; so I asked it here why is that so? $\endgroup$ – user231343 Aug 1 '15 at 3:31
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Michael Hardy has already explained why one wants to say $0\cdot(+\infty)$ is undefined (or indeterminate) in some contexts and 0 in at least one other context, namely integration. I think the value 0 is appropriate in geometric contexts too. Imagine a "rectangle" of 0 height but infinite width; it's essentially just a line, and its area, which ought to be length times width, is 0. The value 0 also makes sense in some combinatorial contexts. Think of $a\cdot b$ as the number of ordered pairs you can form with the first component taken from a given set of size $a$ and the second component taken from a given set of size $b$. If the first set is empty ($a=0$), then there are no such ordered pairs, even if the second set is infinite. (Admittedly, in such a situation, I'd rather use an infinite cardinal for the size of the second set, rather than just $+\infty$.)

I'll go out on a limb and claim that 0 is always the right answer except when you're dealing with a limiting process, with a factor approaching 0 and another factor approaching $+\infty$, as in the first part of Michael Hardy's answer. In other words, the formula $0\cdot(+\infty)=0$ works fine as long as you don't get the idea that the multiplication operation is continuous at $(0,+\infty)$.

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  • $\begingroup$ The "rectangle" you said but with 0 width and infinite height is nonzero (like Dirac delta function). - and the line means $(-\infty, +\infty)$ not $[-\infty, +\infty]$ $\endgroup$ – user231343 Aug 1 '15 at 4:21
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    $\begingroup$ @Edi No, distributions (like the delta function) are quite different from what I was talking about. The relevant rectangle (oriented to match your delta function picture) is $\{(0,y):y\geq 0\}$, and its area is 0. The attempt to describe the delta function as being "$0$ everywhere except $+\infty$ at $0$" gives a very inaccurate picture of it, for precisely this reason. $\endgroup$ – Andreas Blass Aug 1 '15 at 4:25
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If $f(x)\to 0$ and $g(x)\to\infty$ as $x\to\text{something}$, then $f(x)g(x)$ could approach any number or $+\infty$ or $-\infty$, depending on what functions $f$ and $g$ are. For example, consider $\vphantom{\dfrac\int\int}x\cdot \dfrac 6 x$ as $x\to 0$.

There is at least one context in which it makes sense to say $0\cdot\infty=0$, and that is the theory of integration: $$ \int_{-\infty}^\infty 0\,dx = 0, $$ and if $\displaystyle f(x)=\left.\begin{cases} \text{some bounded function} & \text{if } 0\le x\le 5 \text{ or }5<x\le 8, \\ +\infty & \text{if }x=5, \end{cases}\right\}$, then $\displaystyle \int_0^8 f(x)\,dx = {}$whatever number it would have been if we had put any number at all in place of $+\infty$ in the piecewise definition.

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  • $\begingroup$ In $\int_{-\infty}^\infty 0\,dx = 0$, it means $0\times 2\times \infty$ in which $\infty \in ...,\infty)$. But in Dirac delta function, $\infty$ in only one point makes integral nonzero. $\endgroup$ – user231343 Aug 1 '15 at 4:18
  • $\begingroup$ @Edi : True, except that nobody would actually say that the value of the delta function at some point is $\infty$ except when speaking of intuitions. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 1 '15 at 6:29

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