I am reading "Principles of Mathematical Analysis" by Walter Rudin.
Definition 4.1
Let $X$ and $Y$ be metric spaces; suppose $E \subset X$, $f$ maps $E$ into $Y$, and $p$ is a limit point of $E$. We write $f(x) \to q$ as $x \to p$, or $$\lim_{x \to p} f(x) = q$$
if there is a point $q \in Y$ with the following property: For every $\epsilon > 0$ there exists a $\delta > 0$ such that $$d_Y(f(x), q) < \epsilon$$
for all points $x \in E$ for which $$0 < d_X(x, p) < \delta.$$
When $p$ is an isolated point of $E$, $$\lim_{x \to p} f(x) = q$$ is not defined.
Let $p$ be an isolated point of $E$.
There exists a $\delta_0 > 0$ such that there are no points $x \in E$ for which $$0 < d_X(x, p) < \delta_0.$$
So, for every $\epsilon > 0$ and for every $q \in Y$ $$d_Y(f(x), q) < \epsilon$$
for all points $x \in E$ for which $$0 < d_X(x, p) < \delta_0.$$
Rudin didn't want multiple limits at $p$?
$p$ is a point in $E$, so I think it is not natural to exclude limit at $p$.
Best Answer
I guess not, since the definition appears to work "vacuously" for every $q\in Y$.
It seems natural to say the definition doesn't work for an isolated point.