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Since $\infty>0$, so $1/\infty >0$, thus I think $1/\infty$ should be infinitesimal, but the calculus book says $$\lim_{x \to \infty} \frac{1}{x}= 0$$

So is $1/\infty$ zero or infinitesimal ?

P.S. I mean are $1/\infty$ and $\lim_{x \to \infty} 1/x$ the same thing here?

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closed as unclear what you're asking by Najib Idrissi, user99914, Watson, N. F. Taussig, Kamil Jarosz Feb 16 '16 at 11:58

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    $\begingroup$ Id don't really understand your question. $\frac{1}{\infty }$ has no meaning mathematically. $\endgroup$ – idm Nov 9 '14 at 13:26
  • $\begingroup$ 1/x as x approaches infinity is slightly different to saying what is 1/infinity. The limit is indeed 0, but the fraction 1/inf doesn't really make sense mathematically, but I guess I would describe it as infinitesimally small if I had to. $\endgroup$ – Kenshin Nov 9 '14 at 13:27
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    $\begingroup$ $\infty$ is not a real number, so the kinds of algebraic manipulations you are doing here don't make sense. $\endgroup$ – Simon S Nov 9 '14 at 13:28
  • $\begingroup$ If you're talking about limits: $\lim_{x\to\infty}\frac1x=0$, because an "infinitesimal" is not a real number ("real" in the mathematical sense and in the normal sense). $\endgroup$ – Akiva Weinberger Nov 9 '14 at 14:19
  • $\begingroup$ @Mew I agree with your opinion, thanks ! $\endgroup$ – iMath Jan 2 '15 at 3:15
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The issue you have here that, talking about real numbers, there is no such thing as $\frac{1}{\infty}$. All you can talk about are limits in the sense of $\lim_{x\to\infty}{\frac{1}{x}}$.

Now if you have a converging sequence $(a_n)$, and for every $n$ we have $a_n>c$ for some constant $a$, this does not mean that $\lim{a_n}:=a>c$, all you get is $a\geq c$.

The thing here is that, if you talk about an infinitesimal number in the sense of "an arbitrarily small positive number", you need to be aware that there is no such number.

For every real number $a>0$ there is an even smaller number $\frac{a}{2}>0$, so the concept of one smallest positive number does not make sense. And, keep in mind, that a limit, if it exists, is a number. Not a set of numbers or some obscure thing with questionable properties, it is a number. So in this case you might argue that, in some way, yes, your limit is what comes closest to an infinitesimal positive number, just that that is $0$.

Another way to look at this is one way of looking at $\mathbb{R}$ - now I don't know if you've ever heard of that, but a formal construction of $\mathbb{R}$ can be done by looking at Cauchy-sequences of rational numbers, and saying two such sequences are equal if their difference converges to $0$. Or, in different words, gets arbitrarily small. So, by the construction of $\mathbb{R}$, any infinitesimally small number (talking about absolutes here) is automatically the same as $0$. (Formally speaking, this is a bit lacking, because, well, in our definition we have sequences of rationals, not reals. But since you can put your sequence of reals between two sequences of rationals, that can be remedied.)

So, long story short: There is a way to look at $\lim_{x\to\infty}{\frac{1}{x}}$ as the smallest possible "positive" real, just that in this case you need to be aware that this number can only be $0$ and as such is not positive anymore. Either that, or you take a good look at what the limit actually means, put $0$ into it and directly prove that $0$ is your limit.

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    $\begingroup$ There is a well-defined arithmetic in the extended real line $\overline{\mathbb{R}}$ and there, $1/(+\infty)=0$. $\endgroup$ – gniourf_gniourf Feb 15 '16 at 10:26
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Your opening comment is correct if one replaces $\infty$ by an infinite hyperreal $H$, say. Indeed if $H>0$ then one also has $\frac{1}{H}>0$ for the infinitesimal $\frac{1}{H}$. This is true by the transfer principle applied to the usual relation $0<x<y\implies 0<\frac{1}{y}<\frac{1}{x}$.

In an infinitesimal-enriched number system, the operator of taking limit as $x\to\infty$ is decomposed into two stages:

(1) evaluating at an infinite value of $x$, say $x=H$; and

(2) taking the standard part, or shadow.

In the example you considered, $\frac{1}{H}$ is infinitesimal, and its shadow is zero. So the answer to your question is, it depends. Before passing to the shadow, it is infinitesimal. After passing to the shadow, it is zero.

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    $\begingroup$ I think this answer can be improved if you make clear from the get-go you're working in a "nonstandard" realm: it can avoid confusing students who might not otherwise know what a "hyperreal" is and would otherwise attempt to understand it in terms of real numbers. $\endgroup$ – GPhys Feb 16 '16 at 8:31
  • $\begingroup$ @GPhys, I refer you to the comment discussion following this answer for an explanation why I think your suggestion is counterproductive. $\endgroup$ – Mikhail Katz Feb 16 '16 at 8:35
  • $\begingroup$ To be clear, I mean something as simple as "This answer uses nonstandard analysis techniques." Just something so learners know where to look when thrown into bizarro world. $\endgroup$ – GPhys Feb 16 '16 at 8:38
  • $\begingroup$ @GPhys if you think Leibniz and infinitesimals are "bizarro" it is going to be hard for me to take your suggestions seriously. $\endgroup$ – Mikhail Katz Feb 16 '16 at 8:39
  • $\begingroup$ I do! It took me several months to become comfortable with Internal Set Theory. $\endgroup$ – GPhys Feb 16 '16 at 8:41

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