Does set $S=(-1)^n$ have any limit points? This set can have only $2$ values $1$ for even $n$ and $-1$ for odd $n$.
Thus, I don't think it should have any limit points being a finite set. But notes from a reputed institute show that both these ($1$ and $-1$) are limit points.
What am I not understanding?
Same case for $S={\cos(n\pi /2)/nEn}$ - it shows $D(s)=\{-1, 0, 1\}$
 A: The term "Limit Point" is ambiguous, and used in two different ways. One group uses it to refer to the elements of the closure of a set, that is, the set of limits that can be formed from sequences that take values from within the set. In this sense, the limit points of the given set are indeed $1$ and $-1$, as constant sequences will converge to these values.
Another group calls "Limit Points" what the first group would call "Accumulation Points", meaning that there is a sequence taken from the set that converges, but one that is barred from using the point itself. So, isolated points are never limit points in this sense, and as this set is made up purely of isolated points, it has no limit points.
Basically, it's two different nomenclatures.
A: You must discriminate between the sequence $s_n=(-1)^n$ and the set $S=\{s_n\mid n\in \Bbb N\}$. Both provide (different) definitions of the term limit point.
A limit point of a sequence is the limit of an appropriate subsequence. $s_n$ has the (constant) subsequence $\hat s_n=s_{2n}=1$, which has the limit $1$. Same goes for $-1$. These are limit points of the sequence $s_n$.
A limit point $s^*$ of a set $S$ is the limit of a sequence in $S\setminus\{s^*\}$. And here you are right. With the usual topology/metric, $1$ and $-1$ are no limit points of $S=\{-1,1\}$ (and there are none at all).
A: Correct me if wrong:
Distinguish between 
1) limit points of a sequence $(x_n)_{n\in \mathbb{N}},$ 
and 
2)limit points of the set 
$S:= ${$x_n| n\in \mathbb{N}$}.
Your example: $x_n = (-1)^n;$
1) $-1,1$ are limit points of the sequence  $ ( x_n)$; since the subsequences $x_{2n+1} ,$ and $ x_{2n}$ converge to $-1,1$ resp.
2) Now consider the set $S:=$ {$-1,1$};
$-1,1$ are not limit points of the set S, 
they are isolated points.
A: A limit point is by definition less than the limit of a sequence. The sequence you referenced does not have a limit, but two limit points.
