Does evaluating hyperreal $f(H)$ boil down to $f(±∞)$ in the standard theory of limits?

Does evaluating $f(H)$ at an infinite hyperreal $H$ when doing calculus in Robinson’s (Keisler’s) framework amount merely to assigning $f(\infty)$ in the standard theory of limits?

This question has its source in a comment exchange following this answer: Concept behind the limit to infinity?

Solutions Collecting From Web of "Does evaluating hyperreal $f(H)$ boil down to $f(±∞)$ in the standard theory of limits?"

The answer is no. First of all, in the system of hyperreals, there isn’t just one infinite hyperreal $H$, but in fact infinitely many (uncountably many in fact, but that’s not important now), and applying $f$ to them can lead to many results.

Secondly, more importantly, $f(\infty)$ and $f(H)$ will often disagree. Consider the simplest example: $f(x)=\frac{1}{x}$. In what you call “standard theory of limits” we will define $f(\infty)=\lim\limits_{x\rightarrow\infty}f(x)=0$, while in system of hyperreals we have $f(H)=\frac{1}{H}$, which is not equal to zero. Instead, it is an infinitesimal hyperreal number.

$\DeclareMathOperator{\st}{st}$

This is a bit too long for a comment, but I’m familiar with the Nelson approach to Nonstandard Analysis, Internal Set Theory (IST), and I believe an answer in this framework easily translates. In IST, instead of creating new sets such as hyperreals, IST adds axioms to set theory that reveal new numbers in existing sets, such as $\mathbb{R}$. In addition, IST introduces the terms “standard” and “nonstandard” as new terms that can be used to describe sets. I hope that by explaining the situation in IST, it is translatable to the other systems or useful to someone who finds this question.

I will assume we have some $f:\mathbb{R}\rightarrow\mathbb{R}$ for simplicity.

First, if $H$ is an unlimited real (guaranteed to exist for us by the Idealization axiom), it’s important to note that $f(H)$ still exists even if $\lim_{x\rightarrow\infty}f(x)$ does not exist.

For example, let $f(x)=\sin x$. $\lim_{x\rightarrow\infty}f(x)$ does not exist, but $f(H)$ is still well defined for all unlimited $H$. For example, let $H=\pi n$ for $n$ any unlimited $n\in\mathbb{N}$, then $f(H)=0$.

However, suppose the limit does exist. Let us suppose $\lim_{x\rightarrow\infty}f(x)=a$ for some standard $a$. If $H$ is an unlimited (positive unlimited, I will let this be implied now) real, it is not true in general that $f(H)=a$. As others have mentioned, for a function like $f(x)=\frac{1}{x}$, $f(H)$ will be an infinitesimal ($f(H)=\frac{1}{H}$). However, some additional facts will be true since the limit exists.

If $H$ is unlimited, then we have $f(H)\simeq a$ where we read $\simeq$ as “infinitely close to”. We say $a\simeq b\iff\forall ^\mathsf{s}c>0,\lvert a-b\rvert <c$. Here, $\forall ^\mathsf{s}$ is “for all standard”.


Nonstandard analysis can be made to rephrase standard analysis limits. Certainly this relates the idea of evaluating functions “at infinity” to limits. We prove the following statements are equivalent for a standard $f(x)$.

  1. $\lim\limits_{x\rightarrow\infty}f(x)=a$
  2. $a$ is standard and $f(x)\simeq a$ for all unlimited $x$

Note that if $a$ is standard, then (2) implies $\st (f(x))=a$ by definition of the standard part. That is, (2) also means the standard part of $f(x)$ is equal to $a$ for all unlimited $x$.


First, we prove $(1)\implies(2)$.

I first note why $a$ must be standard: Since $f(x)$ is standard and converges by a classical formula, its limit must be standard by the transfer axiom (If there is a unique $a$ that satisfies the classical convergence formula for our standard parameters, $a$ must be standard).

$(1)$ tells us $\forall\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$. Let $\varepsilon$ have a standard fixed value, so this is a classical formula of $M$ taking fixed standard values:
$$\forall ^\mathsf{s}\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$$
The transfer axiom of IST tell us that if there exists an $M$ satisfying this formula, there must exist a standard $M$ satisfying it. In this case, $x\geq M$ is automatically satisfied for any unlimited $x$. So, we may write:
$$\forall ^\mathsf{s}\varepsilon >0,\lvert f(x)-a\rvert <\varepsilon\text{ for all unlimited $x$}$$
By our definition of $\simeq$, this means $f(x)\simeq a$ for all unlimited $x$.
$$\tag*{$\square$}$$


Now, we show $(2)\implies(1)$.

We assume that for any unlimited $x$, $f(x)\simeq a$ and $a$ is standard. Suppose we have a standard $\varepsilon >0$, then the following is true
$$\forall ^\mathsf{s}\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$$
This is true because we can pick any unlimited $M$, and the statement is satisfied by our assumption. Now, we have a completely classical formula except for $\varepsilon$. Since this is a classical formula with standard parameters, the transfer axiom tells us this formula is true $\forall ^\mathsf{s}\varepsilon >0$ as soon as it is true for $\forall \varepsilon >0$, so we have $\forall\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$. In other words,
$$\lim_{x\rightarrow\infty}f(x)=a$$
$$\tag*{$\square$}$$


Please comment if I can clarify any IST-specific issues for you that aren’t clear from a different NSA framework.

I agree with @Wojowu that evaluating $f$ at a hyperreal $H$ carries more information than could be contained by $f(\infty)$ in an Archimedean framework, and would like to provide an additional reason.

Uniform continuity is known to be a global property of $f$ in the sense that there is no pointwise notion of uniform continuity in an Archimedean framework. On the other hand, in a hyperreal framework, one can come pretty close, as follows.

It turns out that a real function $f$ is uniformly continuous on $\mathbb{R}$ if and only if the natural extension ${}^{\ast}\!f$ is S-continuous at every point of ${}^{\ast}\mathbb{R}$. Here S-continuity means that every infinitesimal change of the input leads to an infinitesimal change of the output.

Thus, the squaring function on $\mathbb{R}$ is not uniformly continuous because at an infinite point $H$, and infinitesimal change $\alpha=\frac{1}{H}$ of the input leads to a change in the output which is not infinitesimal but rather appreciable:
$$
(H+\alpha)^2-H^2=H^2+2H\alpha+\alpha^2-H^2=2+\alpha^2.
$$
Thus the failure of the squaring function to be uniformly continuous is captured by its failure to be S-continuous at a single infinite point.

The syntax of the Archimedean framework is too poor to express this distinction; certainly one can’t get anywhere here with merely $f(\infty)$.

Cauchy and Robinson defined continuity in similar way procedurally, namely infinitesimal increment always leading to infinitesimal change in the function.

Depending on how broad a class of functions one wishes to encompass, one can develop infinitesimals using stronger or weaker tools (always in the context of the standard ZFC). Thus, Skolem’s arithmetic with infinite numbers is completely constructive, as analyzed in Stillwell’s

Stillwell, John. Concise survey of mathematical logic. J. Austral. Math. Soc. Ser. A 24 (1977), no. 2, 139–161 (see http://www.ams.org/mathscinet-getitem?mr=465766).

Of course, to handle the squaring function we don’t need more than Skolem’s infinities.