5
$\begingroup$

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.

$\endgroup$
19
  • $\begingroup$ 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. $\endgroup$
    – Alexey
    Feb 13, 2021 at 22:06
  • $\begingroup$ A typical first semester calculus course will define the limit only for limit points of the domain of $f$ - so no $\endgroup$ Feb 13, 2021 at 22:10
  • 1
    $\begingroup$ 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. $\endgroup$
    – user825910
    Feb 13, 2021 at 22:11
  • $\begingroup$ 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. $\endgroup$ Feb 13, 2021 at 22:11
  • $\begingroup$ @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. $\endgroup$
    – Alexey
    Feb 13, 2021 at 22:13

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Very good perspective indeed and not quite obvious. $\endgroup$ Feb 13, 2021 at 22:26
  • 1
    $\begingroup$ 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). $\endgroup$
    – Rob Arthan
    Feb 13, 2021 at 22:31
  • $\begingroup$ @RobArthan I suppose this would be the definition of continuity, which is stronger. Let me amend. $\endgroup$
    – pancini
    Feb 13, 2021 at 22:34
  • 2
    $\begingroup$ 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 $2$". $\endgroup$
    – M. Vinay
    Feb 14, 2021 at 7:09
  • 1
    $\begingroup$ @M.Vinay that’s right. $\endgroup$
    – pancini
    Feb 14, 2021 at 7:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .