# limit epsilon-delta definition vs. continuity

Based on the following problem from this source:

$f(x) = \begin{cases}x^2 &, \text{ if } x \text{ rational} \\ x &, \text{ if } x \text{ irrational}\end{cases}$ has $\lim_{x\to 1} f(x) = 1$

and on this math.stackexchange answer, I suspect that limit does not require a "continuous" domain.

But, why the common $\epsilon$-$\delta$ definition of a limit $\lim_{x\to a}f(x) = L$ ($a$ is neither $-\infty$ nor $\infty$) never emphasizes that point? For example, my textbook simply says $x$ without explaining any property about $x$, and therefore, I was wrongfully led to believe that $x$ must come from a "continuous" domain.

I see that Wikipedia's article on precise definition of limit explicitly states that $x \in D \subseteq \mathbb{R}$. But, I think it may also wrongfully lead people to believe that limit $\lim_{x\to a}f(x) = L$ also works when the domain is so discontinuous like the set $\mathbb{Z}$ of integers and $a$ is neither $-\infty$ nor $\infty$.

Now I am wondering: what is the precise definition of limit $\lim_{x\to a}f(x) = L$ ($a$ is neither $-\infty$ nor $\infty$) especially when it comes to being precise about the $x$?

• What do you mean by a "continuous domain"? Continuity is a property of functions, not of sets.
– anon
Commented Oct 25, 2014 at 15:37
• @learnmore indicates that the formal term is "cluster point". So, what I mean is a domain in the sense of the other math.stackexchange answer where between any two of its elements, there is always another element (e.g., the domain of rational numbers has that property whereas the domain of integer numbers has it not). Commented Oct 25, 2014 at 15:40

When we define limit we consider $a$ to be a cluster point of the domain.By a cluster point we mean the following:
If $a$ is a cluster point of $A$ then $(a-\delta,a+\delta)\cap A\setminus \{a\}\neq\phi$
Let $\epsilon >0$ be arbitrary.Then $f$ has $L$ has limit as $x \rightarrow a$ if corresponding to $\epsilon >0 \exists \delta >0$ such that whenever $x\in (a-\delta,a+\delta)\cap A\setminus \{a\}, |f(x)-L|<\epsilon$ .
• Do you really mean $a$ is a cluster point or $a$ is a limit point instead? Commented Oct 26, 2014 at 1:25