Concept behind the limit to infinity?

I can across transfinite numbers and came up with a thought.

What if$$\lim_{x\to\infty}f(x)=f(T)$$where $T$ was a transfinite number?

Generally, in calculus, I have noted that it is two different things, $\lim_{x\to a}f(x)\ne f(a)$. Which makes me wonder about the situation $a=\infty$. Which means it must mean something when we compare $\lim_{x\to\infty}f(x)$ and $f(\infty)$. The two being different, I guessed the above statement about limits to infinite and transfinite numbers.

Because by definition:$$R<T<\infty$$where $R$ is the real numbers.

By definition:$$\lim_{x\to\infty_1^-}x<\infty_1$$Where the subscript indicates the two infinities are equal. In general, the limit from the left side is close, but always less than the actual value. At least, that is my understanding.

But at the same time, a limit to infinite is a limit by which $x$ exceeds real numbers.

So in a sense, $\lim_{x\to\infty}f(x)=f(T)$.

My question is whether or not my "postulation" is correct.

I have also noted that sometimes, on very rare occasions, the following is true:$$\lim_{x\to\infty}f(x)\ne f(\infty)$$

I will leave this up to you guys.

• I'm sorry, but I don't think this is mathematics. I'm voting to close as such. – pjs36 Dec 20 '15 at 0:53
• I guess the sentence "Because by definition $R < T < \infty$ where $R$ is the real numbers" is what got me. It just looks like a bunch of ideas were thrown together, and then it is "left up to [us]" to try make sense of it. I'll withdraw the close vote, but I think there's a lot to be wary of. – pjs36 Dec 20 '15 at 1:13
• What is the domain of your $f$? Since you write $f(T)$, we must assume that the domain contains $T$ as a member. But can we not change the value of $f$ at $T$ in any way we like, without changing $\lim_{x\to\infty}f(x)$? For example with $f: \mathbb{R} \cup \{ \aleph_0 \} \to \mathbb{R}$ defined by $f(x)=42$ for $x<\aleph_0$ and $f(\aleph_0)=43$, what would you conclude? Am I missing something? – Jeppe Stig Nielsen Dec 20 '15 at 1:27
• What does "transfinite number" mean to you exactly? – Eric Wofsey Dec 20 '15 at 1:51
• The short answer is that for purposes of having limits be $f(Z)$ for some "infinite" $Z$ the thing that is usually done is to extend the real line by two additional points, $+\infty$ and $-\infty$, and define $f(\infty)$ to mean $\lim_{x \to \infty} f(x)$ when that exists. Other number systems that include infinite numbers ( things like "nonstandard analysis" and "surreal numbers") are drastically more complicated than that and not so relevant to calculus at the learning stage. @SimpleArt – zyx Dec 20 '15 at 23:08

The idea is a sound one and is the source of the definition of limit of a real function $f$ via the hyperreals. Namely, to find the limit of $f(x)$ as $x$ tends to infinity, we evaluate $f$ at an infinite number, say $H$, obtaining $f(H)$. This is not quite the limit since it is in general a hyperreal number rather than a real number, so what we do is to round off $f(H)$ to the nearest real number (called its shadow or standard part).

Note 1. For transfinite $T$ one cannot in general define what $f(T)$ is, whereas in a hyperreal framework the extension principle is true. This principle says that every real function admits a natural extension to the hypereals, so that evaluation at $H$ is always possible. For details see the textbook Elementary Calculus by Keisler, available online here: https://www.math.wisc.edu/~keisler/calc.html

Note 2. For some historical background and a qualitative discussion see my answer here: https://physics.stackexchange.com/questions/92925/how-to-treat-differentials-and-infinitesimals/224425#224425

Note 3. Leibniz envisioned many orders of infinitesimals: $dx$ but also $dx^2$, etc. Similarly, Leibniz multiplied his infinitesimals by ordinary numbers, so that $2dx$ is also infinitesimal, etc. By taking reciprocals we get an array of infinite numbers. Similarly, the procedures in the hyperreal extension $\mathbb{R}\subseteq{}^{\ast}\mathbb{R}$ involve an array of infinite numbers rather than a single such number $H$. Thus, the limit of $f(x)$ will exist if and only if the hyperreal numbers $f(H)$ for all infinite $H$ have the same shadow.

• The comment was accurate, and is not addressed by the Note. The extension principle that you mention means that if we could uniquely specify an $H$, then $f(H)$ would be unambiguous. But of course we cannot uniquely determine an $H$, so that $f(H)$ is ambiguous. The only thing that does not depend on the choice of $H$ is the standard part, also known as the limit, and the $f(+\infty)$ notation retains just that. This seems a lot simpler and more canonical than introducing nonstandard analysis (at least for the narrow goal of computing and understanding limits). – zyx Dec 22 '15 at 10:38
• One relevance to teaching calculus is: what happens when a student asks for a concrete example of an infinitely large/small number, or of how your proposed limit procedure works in detail? The options appear to be either to dismiss the question by saying "uh, let's just call it $H$", or to use the ultrafilter construction to write down $H$ as a sequence (Keisler's book does this), which is the same as the standard analysis shorthand of defining an infinitesimal to be a sequence tending to $0$, with a huge layer of added mystery and nonconstructiveness. – zyx Dec 30 '15 at 9:26
• Keisler's book uses ultrafilters, and you are teaching Keisler's book and recommending it to the OP. Keisler also uses Robinson NSA with a transfer principle, not nonstandard arithmetic (which, as discussed, suffers from similar problems, but is not our subject of discussion). Correct me if I'm wrong, but I don't think you would ever answer a student in your NSA-based calculus class by talking about definable sequences in first-order Peano Arithmetic. What would you actually say to an actual student wanting a concrete example, and willing to probe your response with further questions? – zyx Dec 30 '15 at 10:13
• "Sanders himself says in the article that Connes' claims have been largely refuted". Sanders cites two of your anti-Connes polemics on the Arxiv as the sole source of his claim that Connes statements about NSA were "mostly refuted". He appears to be wrong in that assessment, because your arguments in the papers were either pure sociology (Connes did equally nonconstructive work) or irrelevant (there is a "definable" model of NSA; this doesn't address his criticisms, which are essentially the same as my comments here). If there were actual refutations that I missed, please let me know. – zyx Dec 30 '15 at 10:30
• "teaching infinitesimal calculus ... not dependent on any nonconstructivities beyond classical mathematics". There is a huge difference between the two. Teaching standard calculus with classical foundations is constructive everywhere in the classical sense of no serious AC, and also strictly constructive (in Bishop's sense) in practice for every use case that a student would ever see. The nonconstructivities are far past the horizon. In nonstandard calculus, deep nonconstructiveness is lurking in every single exercise, unless the question I raised has a good answer. – zyx Dec 31 '15 at 6:54

Adding larger points at infinity does not lead to anything useful beyond the effect of adding $\pm \infty$ to the system. There is no known additional behavior of functions for large and growing finite real $x$ (which is what we want to analyze) that is not captured by adding $\infty$ or $\pm \infty$ to the domain, yet can be captured with additional transfinite infinite points.

Because the ordinals are discrete, there would not be any meaningful extension of $f(x)$ to values beyond $\omega$ (or $\infty$, or whatever the notation would be for the smallest infinite point added to the reals). You could extend $f$ to be constant and equal to $f(\omega)$, or equal to $0$, or define $f(\omega + n)$ to equal $f(n)$ for all positive integers $n$, but none of these extensions would be useful for analysis of $f(x)$.

• I am sorry, but I like the other answer better. It appears to be more focused on my question. Thank you though, I love your answer too! – Simply Beautiful Art Dec 21 '15 at 20:14
• No worries, thanks. The hyperreal approach was actually addressed in comments earlier, under its different (and more common) name, nonstandard analysis. I left another comment explaining this. – zyx Dec 21 '15 at 21:07
• Oh, I didn't know that. See, I can't know these things, so I just ask you guys... <3 – Simply Beautiful Art Dec 21 '15 at 21:08

Here is (at least) one problem with this conception. In the case of something as simple as, say,

$$L = \lim_{x \to \infty} \frac{x^2+1}{2x^2}$$

we should be able to use the conception to determine that $L = 1/2$. And yet, if we plug in any transfinite number for $x$, we find ourselves unable to perform the arithmetic in any useful way to obtain $L = 1/2$. Any transfinite quantity $T$ we plug in yields the indeterminate form

$$L = \frac{T}{T}$$

It doesn't matter so much whether it's "correct" in some Platonic sense as much as whether we can get any use out of it. Without more of a framework behind your conception, it sure seems at first flush that the answer is no, we can't.

• Why should $L=2$? Shouldn't you get the solution $L=\frac12$? And thank you on your input. I'm not really sure about whether or not my idea was correct. It was just a thought. – Simply Beautiful Art Dec 20 '15 at 0:59
• Oops, yes, I should have $L = 1/2$. I changed my example halfway through and didn't make all the changes. I'll go do that now. – Brian Tung Dec 20 '15 at 1:02
• @zyx, is there a theory of Conway's surreals in nonzero characteristic? Do you have a reference? – Mikhail Katz Dec 22 '15 at 8:41
• Well, there is both characteristic 0 and 2 in Conway's theory, surreals and nimbers. You are right that characteristic is not the issue, and I will repost the comment in a different form. @user72694 – zyx Dec 22 '15 at 10:45
• I don't the line of thinking in the answer invalidates the idea of the question. It leads to Robinson-like infinite numbers, for which $f(T)$ would be defined, but there would arise issues of which of the possible infinite values of $T$ gives the answer. The OP is asking, I think, whether pushing (ordinal) $T$ further and further to the transfinite one can somehow go to the ultimate limit directly. This runs into the difficulty that there is no maximum ordinal, and also the problems of how to extend $f$ discussed in other answers. [This is a reposted edit of an earlier comment.] – zyx Dec 22 '15 at 10:50