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Let's suppose we have a function defined at a single point $f : \left\{ 1 \right\} \to \left\{ 1 \right\}$ defined by $f(x) = x$. Its graph is, therefore, composed of a single point $(1,1)$.

Does the following limit exist?

$$\lim_{x \to 1} \ f(x)$$

I've read this post but it addresses the question in a topology perspective. I'd like to answer this considering a typical first semester Calculus course.

Any help is highly appreciated.

bru1987
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  • It depends on your definition of a limit. The definition i've learnt as a student would imply that you cannot talk of a limit in this case. However, now i believe that the appropriate definition is the one used by Bourbaki (Functions of a real variable), and according to it the limit exists. – Alexey Feb 13 '21 at 22:06
  • A typical first semester calculus course will define the limit only for limit points of the domain of $f$ - so no – Hagen von Eitzen Feb 13 '21 at 22:10
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    Well the definition you would find in a calculus course will say a function is continuous at a point if for all $\epsilon >0$ there is some $\delta >0$ so that for all $y$ such that $0<d(y,x)<\delta$ we have $d(f(y),f(x))<\epsilon$ so this question is inherently topological. From the viewpoint of introductory calculus that limit is not defined, as we are concerned with points near but not equal to $x$, which in this case do not exist. –  Feb 13 '21 at 22:11
  • I don't think you'll find any kind of definitive way this would be treated, but my guess is that most calculus books only define the limit operation when the domain point belongs to the interior of the function's domain (or perhaps belongs to a one-sided neighborhood, to allow for discussion of one-sided limits), and so the issue doesn't arise. – Dave L. Renfro Feb 13 '21 at 22:11
  • @HagenvonEitzen, "A typical first semester calculus course will define the limit only for limit points ..." -- too bad for that typical course. This complicates the matter for no benefit. – Alexey Feb 13 '21 at 22:13
  • @Alexey most first semester calculus courses don't even treat limits rigorously. They are really just used as a pedagogical tool to build intuition for the definition of the derivative. – pancini Feb 13 '21 at 22:15
  • @ElliotG, (1) "most first semester calculus courses don't even treat limits rigorously" -- i did not claim the opposite, (2) using the Bourbaki's definition of a limit would not hurt the definition of the derivative, (3) i have no opinion on using limits as a "pedagogical tool" -- i do not understand what this means... – Alexey Feb 13 '21 at 22:20
  • @Alexey: if it's not too much trouble, can you quote the Bourbaki definition please. – Rob Arthan Feb 13 '21 at 22:40
  • @RobArthan, i think it is obvious: it uses filters, so a limit is defined over a filter in general. For the common notion of a topological limit, the filter of neighborhoods is used. Anyway, it does not matter if this is the way of Bourbaki or of someone else. This is just less trouble, it simplifies proofs, and allows to prove things like $\lim_{\cos x\to 1}\sin x = 0$. The other notion of the limit can be easily recovered as $\lim_{x\to a, x\ne a}$. You may contact me by mail or by chat for more details on Bourbaki's book, if needed. – Alexey Feb 13 '21 at 23:48
  • Thanks, but I asked you to quote the definition rather than write an essay about its merits. It doesn't make sense to say a definition is obvious. As an example, can you give the definition of the English word "nesh" without referring to a dictionary? (It's obvious to me, but that's because I was brought up in a part of England where it's a common word.) – Rob Arthan Feb 13 '21 at 23:52
  • @RobArthan, this is not how mathematical definitions work. They do not define some arbitrary "real life" objects, there should better be a good reason to define a mathematical object the way it is defined. Ok, i am sorry, the book where the (obvious) definition is is not Functions of a real variable but General topology. – Alexey Feb 14 '21 at 00:07
  • @RobArthan, first, a point $a$ is called a limit of a filter $F$ if and only if the filter $F$ is at least as fine as the filter of neigbourhoods of $a$. Then, a point $b$ is called a limit of a function $f$ at $a$ if and only if it is a limit of the image of the filter of neigbourhoods of $a$ by $f$. There may be some details to fill in in the case where $f$ is not defined at $a$... – Alexey Feb 14 '21 at 00:09
  • I am just asking you to quote the text of a definition in a book that I do not have to hand. What do you mean by "that is not how mathematical definitions work"? Just look at the number of different mathematical concepts that are filed under labels like "normal" or "regular" or "special" etc. – Rob Arthan Feb 14 '21 at 00:11
  • ... and (PS) "nesh" is a quality and not a real-life object and I now hope, one day, to find a use for it as a mathematical term (I'm amazed physicists haven't grabbed it already.) – Rob Arthan Feb 14 '21 at 00:13
  • @RobArthan, please contact me by some other means. The book i have is in French (can i just copy and paste?), and the definition is split in many parts due to its generality, it would not fit into comments. But anyone can come up with this definition after thinking for a while. – Alexey Feb 14 '21 at 00:14
  • I think it's probably the same as the definition via filters in Kelley, which I'll look at tomorrow. My French is bad, but probably good enough to cope with the Bourbaki quote. But don't you see that it is useless to say "anyone can come up with this definition" (of a concept) unless there is an argument to say that it is the only possible defintion (of that concept). – Rob Arthan Feb 14 '21 at 00:17
  • @RobArthan, this is not a quote, but this is what it boils down to: $b$ is a limit of $f$ at $a$ if and only if $(\forall\epsilon > 0)(\exists\delta > 0)(\forall x\in D_f)(|x - a| <\delta\Rightarrow |f(x) - b| <\epsilon)$. – Alexey Feb 14 '21 at 00:27
  • Let us continue this discussion in chat. – Alexey Feb 14 '21 at 00:33
  • Hint: not necessarily, because for limits you don't consider the point, if the values that the function can assume aren't converging for the prescribed limit. – João Víctor Melo Feb 13 '21 at 22:20

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It depends on the definition, but here is one answer.

You can definite limits in a very general setting: for any topological space $X$, we say that a sequence $\{x_n\}$ converges to $x\in X$ if, for any open set $U\subset X$ containing $x$, there exists $N\in\Bbb N$ for which $n>N$ implies $x_n\in U$.

For a function $f\colon X\to Y$, we can then define $\lim\limits_{x\to c}f(x)=y$ to mean "if $\{x_n\}$ is a sequence of points in $X\setminus\{c\}$ which converges to $c$, then $\{f(x_n)\}$ converges to $y$."

So for the function $f\colon \{1\}\to\{1\}$, there are no sequences at all with values in $\{1\}\setminus\{1\}$. So it is vacuously true that, for every such sequence $\{x_n\}$, the corresponding $\{f(x_n)\}$ converges to $1$.

In my first answer, I really was writing the definition of the statement "$f$ is continuous at $1$ and $f(1)=1$." As in the familiar setting of real functions, continuity implies the existence of the limit.

pancini
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  • Very good perspective indeed and not quite obvious. – João Víctor Melo Feb 13 '21 at 22:26
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    I don't think this is the right perspective. The question is about the limit of a function on the reals and not about the limit of a sequence. The standard definition of "$f(x) \to L$ as $x \to c$" is a statement that only depends on values of $f(x)$ for $x \neq c$ (and is, somewhat bizarrely, true for any $L$ if the domain of $f$ is a singleton). – Rob Arthan Feb 13 '21 at 22:31
  • @RobArthan I suppose this would be the definition of continuity, which is stronger. Let me amend. – pancini Feb 13 '21 at 22:34
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    Suppose the codomain was ${1, 2}$ (but the domain is still ${1}$). Then according to the sequential definition, the limit could be taken as $2$ as well? Because again it would be vacuously true that "for any sequence ${x_n}$ of points in ${1} \setminus {1}$ that converges to $1$, the sequence ${f(x_n)}$ converges to $2$". – M. Vinay Feb 14 '21 at 07:09
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    @M.Vinay that’s right. – pancini Feb 14 '21 at 07:20
  • @ElliotG Thank you, I wasn't sure. – M. Vinay Feb 15 '21 at 01:55