The limit point in the definition of a limit of function Let $E\subseteq \mathbb{R}$ and $f:E\to \mathbb{R}$ be a function. When we define the limit of $f(x)$ as $x\to x_0$ we require $x_0$ to be the limit point of $E$.
Why do we require $x_0$ to be the limit point of $E$?
The definition says: We say that $A=\lim \limits_{x\to x_0} f(x)$ iff $\forall \varepsilon>0$ $\exists \delta=\delta(\varepsilon)>0$ : $\forall x\in E$ with $0<|x-x_0|<\delta$ $\Rightarrow$ $|f(x)-A|<\varepsilon$.
My thoughts: If $x_0$ is the limit of point of $E$ then $\{x\in E:0<|x-x_0|<\delta\}\neq \varnothing$ for any $\delta>0$. Probably this is one of the reasons but I may be wrong!
But what if $x_0$ is not a limit of $E$? For example, let's take some function $f(x)$ defined on $\{0\}\cup (1,2)$, where $x_0=0$.
What difficulties do we have?
Would be thankful if someone provide detailed answer with examples.
 A: The def'n is flawed. It implies that if $x_0$ is not a limit point of $E$ then $$A=\lim_{x\to x_0}f(x)$$ is true for any and all $A,$ e.g. $0=\lim_{x\to x_0}f(x)=1.$
If $S$ is a sentence and $\forall y\; (T)\;$ (or, respectively $\exists y\; (T)\;$) occurs in $S$, where $(T)$ is the rest of the sentence, then in the negation of $S$  this part will change to $\exists y (\neg T)\;$ (respectively $\forall y\;(\neg T)\;$).
So the def'n says $$A\ne \lim_{x\to x_0} f(x) \text { iff } \exists e>0\;\forall d>0\; \exists x\, (0<|x-x_0|<d \land |f(x)-f(x_0)\ge e).$$ Now if $0<|x-x_0|<d$ then $f(x)$ does not exist unless $x\in E,$ so a consequence of the def'n is
$ A\ne \lim_{x\to x_0}f(x)\implies x_0$ is a limit point of $E,$
equivalently,
if $x_0$ is not a limit point of $E$ then $A=\lim_{x\to x_0}f(x).$
The def'n should be amended by inserting "$x_0$ is a limit point of $E$ and" just after the bold-face "iff".
A: I like to work with the following definition of the limit of a function between metric spaces:

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces such that $f:X\to Y$ is a function. Consider as well that we are given a set $E\subseteq X$ such that $x_{0}\in X$ is an adherent point of $E$ and $L\in Y$. Then we say that $f$ converges to $L$ as $x$ approaches $x_{0}$ along $E$ iff for every $\varepsilon > 0$, there corresponds a $\delta > 0$ such that, for every $x\in E$,
\begin{align*}
d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),L) <\varepsilon
\end{align*}

It is slightly different from your definition, because we demand that $d_{X}(x,x_{0}) < \delta$ instead of $0 < d_{X}(x,x_{0}) < \varepsilon$. However your definition can be considered a particular case from the definition I have mentioned. Indeed, it suffices to consider $E\backslash\{x_{0}\}$ and both of them are the same. The set $E$ tells us how we are approaching $x_{0}$.
Still, the answer to your question has not been given. More precisely, why do we need to consider $x_{0}\in X$ to be an adherent point of $E$? In order to answer it, remember that the set of adherent points of $E$ equals $\overline{E} = \text{int}(E)\cup\partial(E)$. Consequently, if $x\not\in\overline{E}$, then $x\in\text{ext}(E)$. This means there exists a positive real number $r > 0$ such that $B(x_{0},r)\cap E = \varnothing$. In other words, for a $\delta$ small enough $(\delta \leq r)$, there is no $x\in E$ such that $d_{X}(x,x_{0}) < \delta$, which makes the definition of limit vacuous.
That's why we require $x_{0}\in X$ to be an adherent point of $E$.
A: Probably, Heine's definition (that is equivalent to epsilon-delta definition) of the limit of a function may answer your question:

For functions on the real line, one way to define the limit of a function is in terms of the limit of sequences. In this setting:
$$\lim _{x\to x_0}f(x)=A$$
if and only if for all sequences $x_{n}$ (with $x_{n} \not = x_0$, $\forall n$) converging to $x_0$ the sequence $f(x_n)$ converges to $A$.

Therefore, if $x_0$ would not be a limit point of $\mathcal{D}(f)$, then how would such $x_n$ exist?
