Do limits leave residual infinitesimals, or do they resolve exactly?

Question

This is honestly a very basic question regarding limits taken to infinity, but I'm curious about a rigorous explanation that doesn't rely upon notational tricks which obscure the operations. I think the easiest way to state the question is to consider,

$$\lim_{n\rightarrow\infty} \frac{1}{n} = 0$$

This limit clearly evaluates to 0 but it has been my unstructured understanding that, strictly speaking,

$$\lim_{n\rightarrow\infty} \frac{1}{n} \not\equiv 0$$

That is, the limit is not defined to be 0 but rather approaches zero; in that there exists no real number closer to zero than the evaluation of this limit. In the same sense that $\sup (a,b)=b$ despite $b\not\in(a,b)$. So, strictly speaking (though it may be an abuse of notation), it would be more correct to write:

$$\lim_{n\rightarrow\infty} \frac{1}{n} \equiv dn \approx 0$$

Where the limit is defined to evaluate to an infinitesimal in $n$, which, outside specific scenarios usually involving derivatives or infinite sums, may be approximated as zero without loss of generality (since carrying the infinitesimals won't change the analysis). In that sense, whether or not to estimate with $dn=0$ is somewhat an axiomatic preference or convention which depends upon the context of the problem.

Is this the correct way to view limits, or is there some manner by which the limit evaluates precisely to 0, without leaving a residual infinitesimal?

I have seen several explanations, but the internet is inconsistent and my old college calculus I textbook stuck with the high-level evaluation of limits without digging too much into infinitesimals.

Answer 1

A key point is that the sentence "the limit approaches zero" is not properly phrased in the first place.

This is because the kind of object that can approach a number is a sequence, not a limit. The "limit of a sequence" is then defined to be the number that that sequence approaches.

Notice that a limit is defined as a number and as such, like any other number, might or might not equal zero, but it makes no sense to talk about it approaching zero or having some kind of infinitesimal value.

In terms of notation, the statements "the sequence $1/n$ approaches $0$" and "the limit $\lim_{n\rightarrow \infty} 1/n$ equals $0$" both make sense and mean the same thing. On the other hand, the statement "the limit $\lim_{n\rightarrow \infty} 1/n$ approaches $0$" is not meaningful as the verb "approaches" describes something that only a sequence can do, and so cannot apply to the number $\lim_{n\rightarrow \infty} 1/n$,.

Answer 2

That is, the limit is not defined to be 0 but rather approaches zero

No. The value of $\frac{1}{n}$ approaches zero but never reaches it.

But the limit is actually (by definition) the number that is approached, even if it is never reached.

Answer 3

An infinitesimal number (or quantity) $\epsilon$ can be thought of as the reciprocal (i.e., inverse) of an unlimited number $H$, in other words $\epsilon=\frac1{H}$. Leibniz thought of both such $\epsilon$ and $H$ as inassignable numbers. For an analysis of Leibnizian infinitesimals, see this recent publication:

Bair, J.; Błaszczyk, P.; Ely, R.; Katz, M.; Kuhlemann, K. "Procedures of Leibnizian infinitesimal calculus: An account in three modern frameworks." British Journal for the History of Mathematics 36 (2021), no. 3, 170-209. https://doi.org/10.1080/26375451.2020.1851120, https://arxiv.org/abs/2011.12628

Leibniz's distinction between assignable and inassignable numbers was formalized in Abraham Robinson's approach to infinitesimal analysis in terms of the distinction between standard and nonstandard numbers.

As far as the definition of infinitesimals: Leibniz defined an infinitesimal as a number smaller than every assignable (positive) number. In my teaching experience, students find this "easy to grasp". The definition in nonstandard analysis is similar: a positive infinitesimal is a number smaller than every standard positive number.

For every limited number $x$, its standard part or shadow (denoted "sh") is (as the name suggests) a standard number $r$, in other words $\text{sh}(x)=r$. The standard part of an infinitesimal is zero: sh$(\epsilon)=0$.

With these preliminaries, we can now say that the limit of a convergent sequence $(x_n)$ is the standard part of $x_H$ for an unlimited $H$: $$ \lim_{x\to\infty} x_n=\text{sh}(x_H^{\phantom{I}}). $$ For example, for the sequence $x_n=\frac1n$ we obtain that for an unlimited $H$, the term $\frac{1}{H}$ is infinitesimal and therefore its standard part is $0$ "on the nose". Thus, "residual infinitesimals" are indeed involved in evaluating a limit, but they disappear once we apply the standard part.

For an elementary axiomatic approach to infinitesimal analysis (that does not involve either the axiom of choice or ultrafilters), see this introduction.

Answer 4

In this language, limits project from mixed finite and infinitesimal expressions to finite expressions, eliminating the infinitesimals. The result of evaluating a limit (for calculus of a real variable) is a real number, not an infinitesimal. (In the language of nonstandard analysis, limits give "standard parts" (real numbers). Caution: Do not dive down the nonstandard analysis rabbit hole until you have had a course in analysis and possibly one in abstract algebra.)

In a "typical" calculus class, you should not be thinking that infinitesimals are objects you calculate with -- they can be used to estimate, to approximate, and as mental modelling aids. They are neither quantities nor numbers and should not appear where quantities or numbers should appear.

Answer 5

The limit itself is exact. It is only the value of the function you are limiting that tends to the limit value (or is possibly equal to it). This can be realised by analysing the $\epsilon$-$\delta$ definition of a limit. The definition is as follows: $$\lim_{x\rightarrow c} f(x) =L \iff \forall \, \epsilon > 0, \exists \, \delta > 0 \; s.t. \;0<|x - c| < \delta \longrightarrow |f(x) - L| < \epsilon$$ Or in words: The functional value must get closer to the limit value as the x value gets closer (but never equal to) c.

Looking at the last part ($|f(x) - L| < \epsilon$) you can see that it is the functional value that lies in the interval $(L-\epsilon,L+\epsilon)$ ,i.e., it can be equal to the limit or not equal to the limit in the interval $(c-\delta,c)\cup(c,c+\delta)$. That is, it may be equal to (or not) in the interval excluding the value at which the limit is evaluated.

Thus in your example $$\lim_{n\rightarrow\infty} \frac{1}{n}$$ the value is 0 not an infinitesimal that tends to 0.

Answer 6

A sequence is something that may (or may not) approach a number. For example, the sequence $1/n$ for $n \in \mathbb{N}$ approaches the number $0$, in the sense that the sequence gets arbitrarily close to $0$ if you go far enough in the sequence.

The limit operation $\lim$ is something which takes a sequence and produces a real number. Namely, if the sequence approaches a real number $r$, the limit of this sequence is the number $r$.

In symbols: a sequence has type $\mathbb{N} \to \mathbb{R}$, the limit operator has type $(\mathbb{N} \to \mathbb{R}) \to \mathbb{R}$, i.e. it takes a sequence as an argument and output a real number, and the limit of a particular sequence has type $\mathbb{R}$, i.e. it is a real number.

It's just a matter of keeping the types of different objects straight. The limit, if it exists, outputs an object of the type real number. Of course, it makes zero sense to say that one real number approaches another. The number $1/10$ does not "approach" the number $0$, the number $\sqrt{2}$ does not "approach" the number $-5$ etc.

You use the following example:

In the same sense that $\sup(a,b)=b$ despite $b \notin (a,b)$

Yes, this is precisely the same principle: $\sup X$ for a bounded set $X$ of real numbers is a real number, just like $\lim_{n \to \infty} 1/n$. You don't say that $\sup (0,1)$ "approaches" $1$ or that although $\sup (0,1) = 1$, there is some other sense of equality in which "strictly speaking" $\sup (0,1) \neq 1$. Mathematics does not distinguish between two categories of equality: those which hold but not "strictly speaking" and those which "hold strictly speaking". There is just equality and that's it.

Note also that why it is possible to build a theory of calculus which does genuinely admit infinitesimals, this simply a different theory than the ordinary calculus that the vast majority of people are taught in Calculus classes. It's not a more precise theory, it's just a different theory. In particular, the fact that you can set up such a theory with infinitesimals in no way implies that the ordinary theory is somehow imprecise or that it "strictly speaking" involves infinitesimals which are then ignored for "practical purposes" but "strictly speaking" should be there.

Ordinary calculus should not at all be thought of as involving some sort of imprecision regarding infinitely small quantities. On the contrary, ordinary calculus is extremely clear and precise about the matter: it states that no infinitely small quanties exist. It does not at all state that infinitely small quantities exist but can be "safely ignored for most purposes" or whatever.