# Is “taking a limit” a function? Is it a procedure? A ternary operation?

I was sitting in analysis yesterday and, naturally, we took the limit of some expression. It occurred to me that "taking the limit" of some expression abides the rules of a linear transformation $$\lim_{x \rightarrow k}\ c(f(x)+g(x)) = c \lim_{x \rightarrow k} f(x) + c\ \lim_{x \rightarrow k} g(x),$$ and (my group theory is virtually non existent) appears also to be a homomorphism: $$\lim_{x \rightarrow k} (fg)(x) = \lim_{x \rightarrow k} f(x)g(x),$$ etc.

Anyway, my real question is, what mathematical construct is the limit?

• I would avoid defining $\lim$ as a function/relation/etc. but rather $\lim_{x\to k} f(x)=a$ as a notational convention or abbreviation for a much more complicated statement. You might as well ask whether $\dim$ is a function from vector spaces to natural numbers, but you will have trouble defining the "set of all vector spaces". Instead writing $\dim(V)=n$ is an abbreviation of a much longer statement about bases, linear indendence, etc. – M. Winter Feb 2 at 20:13
• This is exactly what I'm wondering. Of course I understand that it is a convention of notation, but it's not only a convention of notation. It can't be "turtles all the way down". There has to be some foundational notion at the bottom of the abstraction pile which lends itself to a thorough understanding of what's happening when one takes a limit. That was my question. @M.Winter – solidstatejake Feb 3 at 14:37
• Similar question (old, but it turned up on the front page recently): What kind of "mathematical object" are limits? – Calum Gilhooley Feb 3 at 20:26

In general, let $$X, Y$$ be topological spaces, and $$x_0$$ a non-isolated point of $$X$$. Then strictly speaking, "$$\lim_{x\to x_0} f(x) = L$$" is a relation between functions $$f : X \to Y$$ and points $$L \in Y$$ (the equality notation being misleading in general).

Now, if $$Y$$ is a Hausdorff topological space, it happens that this relation is what is known as a partial function: for any $$f : X \to Y$$, there is at most one $$L \in Y$$ such that $$\lim_{x\to x_0} f(x) = L$$. Now, for any relation $$R \subseteq (X \to Y) \times Y$$ which is a partial function, we can define a corresponding function $$\{ f \in (X \to Y) \mid \exists y \in Y, (f, y) \in R \} \to Y$$ by sending $$f$$ satisfying this condition to the unique $$y$$ with $$(f, y) \in R$$. Then that somewhat justifies the "equality" in the notation $$\lim_{x\to x_0} f(x) = L$$, though you still need to keep in mind that it is a partial function where $$\lim_{x\to x_0} f(x)$$ is not defined for all $$f$$. (This part relates to the answer by José Carlos Santos.)

Building on top of this, in the special case of $$Y = \mathbb{R}$$, we can put a ring structure on $$X \to Y$$ by pointwise addition, pointwise multiplication, etc. Then $$\{ f : X \to \mathbb{R} \mid \exists L \in \mathbb{R}, \lim_{x\to x_0} f(x) = L \}$$ turns out to be a subring of $$X \to \mathbb{R}$$, and the induced function from this subring to $$\mathbb{R}$$ is a ring homomorphism. (More generally, this will work if $$Y$$ is a topological ring. Similarly, if $$Y$$ is a topological vector space, then the set of $$f$$ with a limit at $$x_0$$ is a linear subspace of $$X \to Y$$ and the limit gives a linear transformation; if $$Y$$ is a topological group, you get a subgroup of $$X \to Y$$ and a group homomorphism; and so on.)

• +1, though I don't think your hesitation with respect to the equals sign is justified; $\frac{6}{x}$ is not defined when $x = 0$, but this certainly doesn't stop us from writing $\frac{6}{2} = 3$. – ruakh Feb 2 at 1:34
• Perhaps Daniel is thinking along these lines: quantamagazine.org/… – Lawnmower Man Feb 2 at 5:07
• That comment was in the context of the general case (not necessarily Hausdorff $Y$) in which you do not have uniqueness of the limit. (So, roughly the same objection that many people have to the $f = O(g)$ notation.) – Daniel Schepler Feb 2 at 7:00
• You don't need a space to create Lim, but limits on spaces are just mostly known. – sanaris Feb 2 at 15:29
• @sanaris How would you take a limit without a space? – Jam Feb 2 at 21:50

Let $$I\subset\mathbb R$$ be a subset of $$\mathbb R$$ such that $$k$$ is an accumulation point of $$I$$. Let$$R=\left\{f\colon I\longrightarrow\mathbb R\,\middle|\,\lim_{x\to k}f(x)\text{ exists}\right\}.$$Then $$(R,+,\times)$$ is a ring and the map$$\begin{array}{ccc}R&\longrightarrow&\mathbb R\\f&\mapsto&\lim_{x\to k}f(x)\end{array}$$is a ring homomorphism.

• what if $f$ is unbounded around $k$ or am I missing something? E.g assume $0\in I$, then $\frac{1}{x^2} \xrightarrow[x\to 0]{} \infty$. – Alvin Lepik Feb 1 at 16:49
• Such a function $f$ will not belong to $R$; in the words, the limit $\lim_{x\to k}f(x)$ will not exist then. – José Carlos Santos Feb 1 at 16:54
• Take $I=(0,1)$. Then $f\in R$ and its limit is infinity as $x\to 0$, where $0$ is an accumulation point of $I$. The limit does exist. – Alvin Lepik Feb 1 at 16:55
• No, it doesn't. I was thinking only about the case in which $\lim_{x\to k}f(x)$ exists in $\mathbb R$. – José Carlos Santos Feb 1 at 16:57
• Ah, right, my bad. Would be helpful to say limit exists and is finite or some such. – Alvin Lepik Feb 1 at 16:58

To be more precise, "taking $$\limsup$$ of a sequence" and "taking $$\liminf$$ of a sequence" are both functionals, i.e. linear functions from a vector space whose elements are sequences to its underlying field of real numbers. (You can also consider it a functional in a more advanced way, by putting a topology on the vector space for which the operation is continuous.)

This captures just the additivity. To capture multiplicativity, you need to consider the space of sequences not just as a vector space, but as a ring (as before, you can put a topology on this ring, in which case it becomes natural to consider a space of sequences satisfying enough conditions to become a Banach algebra). Then $$\limsup$$ and $$\liminf$$ become ring homomorphisms from a ring whose elements are sequences to the real numbers.

• Aren't $\limsup/\liminf$ of a sequence different that the limit of a function? – solidstatejake Feb 1 at 16:16
• Not when the limit exists. The reason I mention these two is that they are well-defined for all sequences, whereas $\lim$ is not. In fact, if you try to extend the definition of limit to all sequences, all possible extensions interpolate between liminf (the smallest extension) and limsup (the largest). – pre-kidney Feb 1 at 16:16
• Umm, $\liminf$ and $\limsup$ aren't additive in general - nor does either one respect multiplication by negative scalars. (Plus, you would need to restrict to the space of bounded functions in order to avoid having values of $+\infty$ or $-\infty$.) – Daniel Schepler Feb 1 at 16:23
• @DanielSchepler yes, which is why I was careful to avoid saying that the vector space consists of all sequences in my answer. But thanks for pointing it out explicitly. You do raise a good point though that we are not gaining much by going from lim to limsup/liminf since we will need to consider restrictions either way. – pre-kidney Feb 1 at 16:33