Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is whether by definition there can exist a limit of a function that's defined only in one point ( or in several points but there's no interval in which the function is defined).

This came up when thinking about $\lim_{x \to 0}{\sqrt{-|x|}}$

share|cite|improve this question
Note that there are several ways to interpret your question as posed, and your example gives a slightly different interpretation. – Calvin Lin Dec 29 '12 at 14:16
It is possible to define continuity at a single point, but the definition of $\lim_{x\to a}$ is about points close to, but not equal to, $a$ – Thomas Andrews Dec 29 '12 at 14:28
It is not possible to talk about the existence of a limit since there doesn't any sequence of points {$a_n$}$_{n\ge 1}$ in the domain of the function such that $a_n \ne 0$ and $a_n$ converges to zero, thus it is not possible to check if the image sequence has a limit or not. – Siddharth Joshi Nov 23 '15 at 17:11
up vote 4 down vote accepted

No. For the limit $\lim_{x\to a}f(x)$ to have a meaning $x$ will have to be able to go as close to $a$ as desired. Formally if $f:X\subseteq \mathbb{R}\to \mathbb{R}$ we demand that $a$ is an accumulation point of $X$, that is: $$\forall \epsilon>0\ \exists x\in X:\ 0<\left|x-a\right|<\epsilon$$ which in our case is not true. The point $0$ is isolated in $\left\{0\right\}$ and so the symbol $\lim_{x\to a}f(x)$ has no meaning.

You may want to note that $f$ is however continuous at $0$, in fact any function is continuous at the isolated points of its domain.

EDIT: To be unambiguous here is my definition of limit:

If $f:X\subseteq \mathbb{R}\to \mathbb{R}$ and $a$ is a limit point of $X$ then $\lim_{x\to a}f(x)=L\in \mathbb{R}$ if $$\forall \epsilon>0\exists \delta>0:0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon$$

share|cite|improve this answer
Well this confuses me a little because according to wiki continuity of a function at $ x=a $ requires that $ \lim_{x\to a}f(x) = f(a) $ , but according to what you said there's no limit at all at $ x=a $ let alone $ f(a) $ – Ord Dec 29 '12 at 14:25
@Ord This is true if $a$ is an accumulation point of $X$, then you require that the limit exists and its value is $f(a)$. If it isn't then $f$ is automatically continuous there. But because continuity at isolated points is uninteresting, people usually forget that $a$ must be a limit point of $D_f$ – Nameless Dec 29 '12 at 14:26
I've tried reading the definition all over again but I haven't seen any reference to the point being(or not) an accumulation point in the definition and to tell you the truth the definition at my Uni was the same, so is this definition not complete or am I just missing something ? sorry for asking so many questions I just want grasp the whole concept continuity and limits and at times it can be a bit unclear. So thanks. – Ord Dec 29 '12 at 14:42
@Ord The "accumulation point" is added so that $0<\left|x-a\right|<\delta$ "makes sense", that is there exist $x$ so that it is true. In most definitions this is not explicitely written out, but I choose to do so to emphasise the importance of $a$ being a limit point – Nameless Dec 29 '12 at 14:46

A limit of a real-valued function defined only at one point does not exist, such as wikipedia defines limits. The limit value of a function at a point $c$ is defined from the function values of points arbitrarily close to, but not equal to $c$.

An example would be the function that has $f(0) = 1$ and is $0$ everywhere else. Then the limit value should be defined so that it is $0$ at every point. Thus the definition of limit at $0$ cannot take the function value at $0$ into account.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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