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Just wondering, while studying limit, if $x\over x$ is continuous at $0$. $f(0)={0 \over 0}$ ,, but $x/x=1$. In this case, is it continuous at $0$?

quid
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Kaytlyn
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  • What exactly is your definition of the function $;\frac xx;$ on the real line and, in particular, at $;x=0;$ ? – Timbuc Jan 23 '15 at 19:21
  • @Timbuc, it does not matter how you define or not define the function at $x = 0.$ the limit is $1.$ – abel Jan 23 '15 at 19:28
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    @abel yes, whether the function is continuous however depends on its value at $0$. – quid Jan 23 '15 at 19:30
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    ...what is $p$? – daOnlyBG Jan 23 '15 at 19:42
  • @abel, I think I know that, yet that was not the question. – Timbuc Jan 23 '15 at 19:44
  • the function $x/x$ is not even defined at $x=0$, though it can be uniquely extended to a continuous function on $\mathbb{R}$ by virtue of the fact it only has a removable, isolated singularity. this is the idea behind holomorphic extensions – obataku Jan 23 '15 at 19:49
  • @daOnlyBG given that "p" is close to "0" on various keyboards, as well as other considerations, it should be $0$ in all likelihood. – quid Jan 23 '15 at 19:51
  • Despite the kind of "clueless" nature of this question, I don't think that it deserves the down-votes (and in a certain way, the well elaborated answer below proves it). – barak manos Jan 23 '15 at 20:27

2 Answers2

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The question is not quite precise but:

  • The limit $\lim_{x \to 0} x/x$ exists and is $1$.

  • The function $x \mapsto x/x$ for $x \in \mathbb R \setminus \{0 \}$ is not defined at $0$ and it makes no sense to ask about its continuity there.

  • The function $x \mapsto x/x$ for $x \in \mathbb R \setminus \{0 \}$ can be extended to a function on $\mathbb R$ via defining it to be $1$ for $x=0$. This is then obviously a continuous function. One could also extend it in some other way, in which case it would not be continuous.

user26486
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quid
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    Great answer. I was trying to push the OP towards something like this. +1 – Timbuc Jan 23 '15 at 19:45
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    Might have not up voted if not for "makes no sense to ask about continuity here". Time and again people will make the mistake of saying it is discontinuous in this case. – Git Gud Jan 23 '15 at 19:48
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    @GitGud Then ask "Is the square root continuous at 'shoe'?" – Daniel Fischer Jan 23 '15 at 20:00
  • @GitGud Doesn't that depend on how you use the definitions and terms? The definition I know is, a function $f$ is continuous at $c$ iff $f(c)$ is defined, $\lim_{x\to c}$ exists, and $\lim_{x\to c}=f(c)$. In this case, $f(0)$ is not defined, so $f$ is not continuous at $0$. If "discontinuous at $c$" simply means "not continuous at $c$," then isn't $f$ discontinuous at $0$? On the other hand, the whole function is continuous when it is continuous at every point in the domain, and $f$ is continuous, since $0$ is not in its domain. An easy source of confusion, but is my first statement wrong? – KSmarts Jan 23 '15 at 20:11
  • @DanielFischer To me that question makes no sense. – Git Gud Jan 23 '15 at 20:16
  • @GitGud That's the point. Asking whether a function is continuous at a point outside its domain makes no sense. If you use a sufficiently absurd example, you might get the point across. – Daniel Fischer Jan 23 '15 at 20:17
  • @KSmarts I agree with you. But many times the definition goes like "given $c$ in the domain of $f$ (...)". using this definition, if you're not given $c$ is the domain, there's nothing you can say, it's senseless. – Git Gud Jan 23 '15 at 20:18
  • @DanielFischer Oh, I see. I thought you were disagreeing with me. – Git Gud Jan 23 '15 at 20:19
  • @GitGud It reminds me of a question I asked myself. – drhab Jan 23 '15 at 20:49
  • I think many authors use "discontinuity point" even at points not in the domain of definition but for which there is at last some one-sided non-trivial neighborhood where the function's defined, e.g.: $;\frac{\sin x}x;$ is discontinuous at $;x=0;$ . Yes, $;x=0;$ is not a natural definition point of this function, yet it has a non-trivial neighborhood where the function is defined. Something similar to vertical asymptotes of functions. A matter of (dis)agreement, I think. – Timbuc Jan 23 '15 at 22:11
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A function $f$ is continuous at $c$ iff $f(c)$ exists, and $f(c)=\lim_{x\to c}f(x)$.

In your example, $f(x)=\frac xx$ and $c=0$. Since $f(c)$ does not exist, your function is not continuos at that point.

Akiva Weinberger
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