If $V$ is a neighbourhood of $a$ having $E= D \cap V$ , show that $a \in E'$ I'm having a little problem in understanding this question...

Let $f \colon D \rightarrow \mathbb{R}$ with $D \subset \mathbb{R}$ and $a  \in D'$. If $V$ is a neighbourhood of $a$ and $E= D  \cap V$, show that $a  \in E'$.

The exercise also asks to show that if   $f|_E$ has a limit at the point $a$ then $f$ also has a limit at the point $a$ .
More specifically, I don't understand how to analyze the restricted function...
Thanks the attention!
(This is an exercise of my real analysis list of exercises. I spent a good time trying to do it, but I'm stuck... If anyone could give me a light... I'm studying for my final exam...)
 A: Expanded, given that apparently it was unclear.
Because $a\in D'$, that means that every neighborhood of $a$ must contain points of $D$ other than $a$. You want to prove that every neighborhood $U$ of $a$ must contain points of $E\cap V$ other than $a$ itself; let $U$ be a neighborhood of $a$. Then $U\cap V$ is also a neighborhood of $a$, so we know that $(U\cap V)\cap D-\{a\}$ is not empty. But $(U\cap V)\cap D = U\cap (V\cap D) = U\cap E$. Hence...
For the second part, let us suppose that $f|_E$ has a limit at $a$, say $r$.
This means that for every open set $W$ that contains $r$, there exists an open set $U$ that contains $a$ and such that $f((U\cap E)-\{a\})\subseteq W$.  To show that $f$ has a limit at $a$, we need to show that there exists an open set $\mathscr{O}$ that contains $a$ and such that $f((\mathscr{O}\cap D)-\{a\})\subseteq W$.
Fix $W$, an open set containing $r$, and let $U$ be an open set that contains $a$ and such that $f((U\cap E)-\{a\})\subseteq W$. Now let $\mathscr{O} = U\cap \mathrm{int}(V)$, the interior of $V$. This is an open set that contains $a$ (since $U$ is open and contains $a$, and $V$ is a neighborhood of $a$ so its interior contains $a$). Will that work?
