limit point

Don’t confuse with accumulation point (of sequences).

Motivation

A limit point of $D$ is a point you can approach from within $D$. That is the precondition for function limits: ${\lim }_{x\to x_0}f(x)$ makes sense only when $x_0$ is a limit point of the domain, otherwise no points of $D$ get near $x_0$ and there is nothing to take a limit of. So they mark where continuity and limits can be asked about.

Limit point of a set

Let $D\subseteq \mathbb{R}$. A point $x_0$ is a limit point of $D$ if every punctured epsilon neighbourhood of $x_0$ (the ball with $x_0$ removed) contains a point of $D$:

$$\forall \epsilon >0:(U_{\epsilon }(x_0)\setminus \{x_0\})\cap D\ne \varnothing$$

Equivalently: some sequence $(x_n)$ in $D\setminus \{x_0\}$ has $x_n\to x_0$.

Removing $x_0$ forces the nearby points to be other members of $D$: $x_0$ cannot count as its own witness, so it need not lie in $D$ at all.

The opposite is an isolated point: a point of $D$ with a neighbourhood holding no other point of $D$. Every point of $D$ is one or the other.

The points of $D=\{1,\frac{1}{2},\frac{1}{3},\dots \}$ on the line. They pile up toward $0$: the gap between consecutive points shrinks to zero, so any ball $U_{\epsilon }(0)$, however small, still contains every $1/n$ with $1/n<\epsilon$, i.e. infinitely many. So $0$ is a limit point, though $0\notin D$ (open circle).
A single point has room around it instead: the neighbours of $\frac{1}{3}$, namely $\frac{1}{2}$ and $\frac{1}{4}$, sit a fixed distance away, so a small enough $U_r(\frac{1}{3})$ holds only $\frac{1}{3}$. Every $1/n$ is isolated this way.

Limit points of common sets

$\{1/n:n\in \mathbb{N}\}$: the only limit point is $0$, and $0\notin D$. Each $1/n$ is isolated.
$(0,1)$: limit points are all of $[0,1]$, including the endpoints $0,1$ that aren’t in the set. No point of the interval is isolated.
$\mathbb{Q}\cap [0,1]$: every $x\in [0,1]$ is a limit point, since the rationals are dense. Uncountably many limit points out of a countable set.
Any finite set: no limit points (always a gap to fit a ball into).
$\mathbb{N}$: no limit point in $\mathbb{R}$, every point is isolated. It accumulates only at $+\infty$, which is why the definition is sometimes extended to $\mathbb{R}\cup \{\pm \infty \}$.

A bounded infinite subset of $\mathbb{R}$ always has a limit point (set form of Bolzano-Weierstrass).