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Say there exists a limit $\lim_{x \to x_0}f(x) = L$. Is it necessary that $f$ be defined at the point $x_0$ itself?
Well, what I think of it is that it's OK to be undefined at that point because I guess that won't cease the limit of that function to exist there would it?

Arnav Das
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  • You mean if it is necessary for $L$ to be a value of $f$? – Yes Nov 17 '15 at 15:46
  • Well it's something like if $L$ should be defined at the point $x_0$ ? Wait am re-editing my quest properly – Arnav Das Nov 17 '15 at 15:48
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    $L$ being defined at the point $x_0$ doesn't make much sense. I suppose you mean does $f$ have to be defined at $x_0$? In that case, no, it does not. (Note that even if it were defined at $x_0$ but $f(x_0) \neq L$, it is still feasible for $\lim_{x \to x_0} f(x) = L$. Hope that helps. –  Nov 17 '15 at 15:57
  • I made an [edit] to your post. Please check that I did not unintentionally change the content of the question. – gebruiker Nov 17 '15 at 16:00
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    no. it is not necessary. if $f$ is defined at $x_0$ and $\lim_{x \to x_0}f(x) = L=f_{(x_0)}$, $f$ is continious at $x_0$ – Prabath Hewasundara Nov 17 '15 at 16:07
  • Then simply check out the definition of limit again; we only consider the behavior of $f$ near $x_{0}$. – Yes Nov 17 '15 at 16:29
  • @gebruiker thanks I was wrong in editing that – Arnav Das Nov 17 '15 at 17:11
  • Technically speaking, the function $f(x) = 1$ defined on the domain $\mathbb{R} \setminus { 0 }$ is a counterexample which is not defined at $x = 0$ but clearly has a limit. – CompuChip Nov 17 '15 at 23:15

3 Answers3

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"No", in a particularly strong sense: The very reason limits came into mathematics was to extract numerical values from difference quotients $$ \frac{f(x) - f(x_{0})}{x - x_{0}} $$ (which are, of course, algebraically indeterminate at $x_{0}$), in the limit as $x \to x_{0}$.

The formal definition of $L = \lim(f, x_{0})$ ("For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - x_{0}| < \delta$, then $|f(x) - L| < \varepsilon$") explicitly avoids evaluation of $f$ at $x_{0}$, precisely because $f$ might be undefined at $x_{0}$.

Andrew D. Hwang
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    +1, yeah, that's precisely what I was going to say. That's the whole point of limits, when there is no value (or a value cannot otherwise be calculated, or in strange cases where the value is separated from the nearby values.). – McKay Nov 17 '15 at 17:39
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    It is not "precisely because f might be undefined at $x_0$". It is precisely because we are not interested in $f$ at $x_0$. – Aloizio Macedo Nov 17 '15 at 18:33
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    @AloizioMacedo: If we're going to split hairs, I'd argue my wording is correct from the perspective of Cauchy, writing down the definition for the first time. Your wording may be more natural from the perspective of a modern student learning about limits, but that's because Cauchy's definition is well-accepted. – Andrew D. Hwang Nov 17 '15 at 19:12
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This is not at all necessary. For example: $$ \lim_{x\to 0} \frac {x^2}x =0, $$ but $\left.\dfrac{x^2}{x}\right\vert_{x=0}$ is undefined.

If we do have that $f$ is defined at $a$ and $\lim\limits_{x\to a}f(x)=f(a)$, then $f$ is called continuous at $x=a$.

gebruiker
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$\lim_{x\to x_0} f(x)=L$ for a $f:D\to\mathbb R$ is defined as

For all sequences $(a_n)$ with $\lim_{n\to\infty} a_n = x_0$ we find $\lim_{n\to\infty} f(a_n) = L$.

To guarantee that $L$ is uniquely defined we need at least one sequence $(a_n)$ of the domain such as $\lim_{n\to\infty} a_n = x_0$. Therefore $x_0$ need to be an accumulation point of the domain $D$.

Conclusion: $x_0$ does not need to be in the domain of $f$ but it must be an accumulation point of the domain.

Stephan Kulla
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