limit epsilon-delta definition vs. continuity

Question

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 $x$?

Answer 1

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 $(a-\delta,a+\delta)\cap A\setminus \{a\}\neq\phi$

So the exact definition of a limit is

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 $x\in (a-\delta,a+\delta)\cap A\setminus \{a\}, |f(x)-L|<\epsilon$ .