Can someone please show me one example of a continuous function: $f : X\subset \mathbb{R} \rightarrow \mathbb{R}$ but $\lim_{x\rightarrow a} f(x) \neq f(a)$ with $a \in X$?

The converse is clear for me, that means if $\lim_{x\rightarrow a} = f(a)$ then $f$ is continuous at $a$.

I was trying to look for some definition of $f$ where we'd have $a \in X' \setminus X$...

I'd appreciate some help! Thanks!

  • 1
    $\begingroup$ I'm a little confused - if a function $f$ is continuous at a point, then doesn't $f(a) = \lim_{x \to a} f(x)$? $\endgroup$ – paulinho Jun 23 at 18:42
  • $\begingroup$ @paulinho That is valid iff $a \in X \cap X'$. $\endgroup$ – Bruno Reis Jun 23 at 18:45
  • $\begingroup$ @BrunoReis what is $X'$? $\endgroup$ – ArsenBerk Jun 23 at 18:50
  • $\begingroup$ @ArsenBerk $X'$ is the set of accumulation points of the domain of our function. $\endgroup$ – Bruno Reis Jun 23 at 18:52
  • $\begingroup$ Then how would you define $f(a)$ for $X'\setminus X$? You only define it on $X$ in your question $\endgroup$ – Severin Schraven Jun 23 at 18:56

I think your confusing things here: it $\;f:X\to\Bbb R\;$ is continuous, then for any $\;a\in X\;$ it must be true that

$$\lim_{x\to a\\x\in X}f(x)=f(a)\;$$

This is just part of definition (or of what follows from it, depending on your particular definition of continuity).

The above has nothing to do with the fact that if $\;\{x_n\}\subset X\;$ , then $\;\lim\limits_{n\to\infty} x_n=x\in X\iff x\in X'\;$ , which is perhaps what you're thinking of.

  • $\begingroup$ It makes sense... I'm thinking about that because that was a question in a test, to find that counterexample... So I was thinking that what was written on a test was true, and it seems that it's not. $\endgroup$ – Bruno Reis Jun 23 at 19:03
  • $\begingroup$ Most probably the wording of the question was different and they meant something else, or perhaps that was a mistake. It occurs all the time... $\endgroup$ – DonAntonio Jun 23 at 19:05
  • $\begingroup$ Thank you anyway @DonAntonio. Now what if $X = (a,b) \cup \{c\}$ where $c \notin (a,b)$ ? Then we can't talk about limit in $c$ since $c \notin X'$... $\endgroup$ – Bruno Reis Jun 23 at 19:09
  • $\begingroup$ Well, if we require $\;x\to a\,,\,\,x\in X\;$ then yes: that cannot be. What is important here is that everything must happen within $\;X\;$ , which is the definition domain of $\;f\;$ . $\endgroup$ – DonAntonio Jun 23 at 19:11
  • $\begingroup$ When you mean everything, you mean that necessarily $ a \in X$? $\endgroup$ – Bruno Reis Jun 23 at 19:20

If $f: X=(0,1) \cup \{2\} \to \mathbb{R}$ is defined by $f(x)=\sin(\frac1x)$ for $x \in (0,1)$ and $f(2)=42$ then $f$ is continuous on $X$ (continuity is trivial on isolated points: take $\delta=1$ for any $\epsilon>0$, e.g.) but $\lim_{x \to 2} f(x)$ does not exist as $2 \notin X'$ (but $2 \in X$); and as a bonus $\lim_{x\to 0} f(x)$ does not exist even though we do have $0 \in X'$.

Or take any function defined on $X=\mathbb{Z}$. No limit exists to points of $X$ (or outside) as $X'=\emptyset$.

  • $\begingroup$ This is not what was asked. For the examples to be relevant you should have $\;f(1),\,f(0)\;$ or something of the like...which you cannot as you defined your function not on those points . $\endgroup$ – DonAntonio Jun 23 at 21:58
  • $\begingroup$ @DonAntonio No, $\lim_{x \to 2} f(x)$ does not exist (so is not equal to $f(2)$), even though $2 \in X$, so it's a valid example. $\endgroup$ – Henno Brandsma Jun 23 at 22:00
  • $\begingroup$ No, it's not since there's an open neighborhood of $\;a=2\;$ in which the function isn't even defined ...you cannot take the limit $\;\lim\limits_{x\to 2}f(x)\;$ ! This is the main point (with a twist) that I tried to explain the OP... $\endgroup$ – DonAntonio Jun 23 at 22:02
  • $\begingroup$ @DonAntonio Exactly, that's why the limit is not equal to $f(2)$: it does not exist.. We seem to agree and not agree at the same time. $\endgroup$ – Henno Brandsma Jun 23 at 22:03
  • $\begingroup$ I try to abide by the very basic definitions of these very basic things. For me to be even able to talk about $\;\lim\limits_{x\to a}f(x)\;$ , the function $\;f\;$ must be defined in some open neighborhood of $\;a\;$, otherwise the whole thing is meaningless... $\endgroup$ – DonAntonio Jun 23 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.