The point $1$ is not a limit point of the set, because there is a neighbourhood of $1$ such that the only point in the set in that neighbourhood is $1$. Use, for example, the interval $(0.9,1.1)$.
In fact no point in the set is a limit point of the set. Around the point $\frac{1}{n}$, you can put the neighbourhood $(\frac{1}{n}-\frac{1}{4^n},\frac{1}{n}+\frac{1}{4^n})$ that contains no point of our set other than $\frac{1}{n}$.
To show that $0$ is a limit point of our set, take an open interval $(-\epsilon,\epsilon)$ about $0$, where $\epsilon$ is small (possibly very small) positive. There is an integer $N$ such that $\frac{1}{N}\lt \epsilon$. Then every point $\frac{1}{n}$ in our set with $n\ge N$ is in that neighbourhood.
So every open neighbourhood of $0$ contains a point of our set, indeed infinitely many points of our set.
Much more informally, we can get arbitrarily close to $0$ from within our set.
A limit point of a set may or may not belong to the set. For example, let $S=(0,1)$, that is, all real numbers $x$ such that $0\lt x\lt 1$. The number $1$ is a limit point of $S$, and is not in $S$.
Let $T=[0,1]$. The point $1$ is a limit point of $T$ and is in $T$.
Let's work in the real line, for concreteness. Consider a sequence $(x_n)_{n \geq 1}$. A limit point of the sequence is a limit point of the set $\{x_n \mid n\geq 1\}$. When you say the limit point, it means that the set $\{x_n \mid n\geq 1\}$ has only one limit point, say, $L$. And this $L$ is the element who satisfies the definition we all know and love: $$\forall \ \epsilon > 0, \ \exists \ n_0 \geq 1, {\rm s.t.} \ n > n_0 \implies |x_n - L| < \epsilon.$$
If the set $\{x_n \mid n\geq 1\}$ has more than one limit point, then the sequence $(x_n)_{n \geq 1}$ does not converge.
Best Answer
Let $x_n=(-1)^n$ and $\alpha=1$. Then $S=\{x_n:n\in\Bbb N\}=\{-1,1\}$ is a finite set, so it has no limit points.
The point of this exercise is that cluster points of sequences and limit points of sets aren’t quite the same thing. A point $x$ is a cluster point of the sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ if some subsequence of $\sigma$ converges to $x$; $x$ is a limit point of the set $S=\{x_n:n\in\Bbb N\}$ if every open nbhd of $x$ contains a point of $S\setminus\{x\}$. If $x$ is a limit point of the set $S$, then every open nbhd of $x$ contains infinitely many different points of $S$; if $x$ is a cluster point of $\sigma$, then every open nbhd of $x$ contains infinitely many different terms of $\sigma$, but those terms might all be the same point, as in the example that I gave above.