Concept behind the limit to infinity?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $f(x)$.