Originally Posted by
Mathstud28 Note: You may want to wait for a more senior member to validate this answer.
My interpretation of your question
"If $\displaystyle f:A\longmapsto \mathbb{R}$ with $\displaystyle A\subseteq\mathbb{R}$, and if $\displaystyle a$ is not a limit point of $\displaystyle A$ then does $\displaystyle \lim_{x\to a}f(x)$ make sense?"
Answer to that question: No. For $\displaystyle \lim_{x\to a}f(x)$ to exist there must be a point $\displaystyle q\in\mathbb{R}$ such that for every $\displaystyle \varepsilon>0$ there exists a $\displaystyle \delta>0$ such that $\displaystyle d_{A}(x,a)<\delta\implies d_{\mathbb{R}}(f(x),q)=|f(x)-q|<\varepsilon\quad x\in A$. But now suppose that $\displaystyle a$ is not a limit point of $\displaystyle A$ then we can pick an open neighborhood of $\displaystyle a$ lets call it $\displaystyle N_{\delta_1}(a)$ such that there does not exist an element of $\displaystyle A$ in that neighborhood. Now choose an value of $\displaystyle \varepsilon$ that requires $\displaystyle \delta<\delta_1$ and we run into a contradiction.
EDIT: I am sorry, I completely for some reason disregarded the $\displaystyle A\subseteq\mathbb{R}$ part. This can be made slightly simpler
By the defintion of limits in $\displaystyle \mathbb{R}\mapsto\mathbb{R}$ for $\displaystyle \lim_{x\to a}f(x)$ to exist there must exist a point $\displaystyle q\in\mathbb{R}$ such that for every $\displaystyle \varepsilon>0$ there exsits a $\displaystyle \delta>0$ such that $\displaystyle |x-a|<\delta\implies|f(x)-q|<\varepsilon~~x\in A$. But now suppose that $\displaystyle a$ is not a limit point of $\displaystyle A$ then there exists a neighborhood $\displaystyle (x-\delta_1,x+\delta_1)$ such that there does not exist an element of $\displaystyle A$ within that neighborhood. Now if we choose an $\displaystyle \varepsilon$ that requires $\displaystyle \delta<\delta_1$ we run into a contradiction.