$0,1,0,1,0,1$... has only $2$ limit points 
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*Prove that the sequence $0,1,0,1,0,1$... has only $2$ limit points : $0$ and $1$.



To be frank, I know the solution to the above particular problem. What I am interested is in knowing a general theorem/proposition (with proof) that can help solve problems like the above one and the ones mentioned below.



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*The sequence $-1,0,1,-1,0,1,-1,0,1$... has only $3$ limit points : $-1$,$0$, and $1$.

*The sequence $1,0,-1,2,0,-2,3,0,-1,4,0,-1$... has only $2$ limit points : $0$ and $-1$ 

*The sequence $1,1,1,1/2,1,1/3,1,1/4,1,1/5$... has only $2$ limit points: $0$ and $1$


I was hoping that one of you could tell me a general property (with proof)  regarding sequences that help in solving the above mentioned problems. 
 A: There is no general formula for finding the limit points of a given sequence. To complete this task you need to consider for each $a \in \Bbb{R}$ the family of neighbourhoods $(a-\epsilon, a+ \epsilon)$. This is an infinite task.
So we need to see the sequence and guess it's limit point, or more generally to provide a criterion where the limit points are unique. For instance you could check if your sequence satisfies the Cauchy property
A: The princple you probably want to use is the following:


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*If you interleave a finite number of infinite sequences, the limit points of the resulting sequence are (the union of) the limit points of the original sequences.


For example, the limit points of $0,1,0,1,0,1,\ldots$ are the limit points of either $0,0,0,\ldots$ or $1,1,1,\ldots$. Both of those are convergent sequences, and their limits are $0$ and $1$.
The limit points of $1,0,-1,2,0,-1,3,0,-1,4,0,-1,\ldots$ are the limit points of $1,2,3,4,\ldots$, $0,0,0,0,\ldots$, and $-1,-1,-1,-1$, which are $\infty$, $0$, and $-1$. Of those, the real limit points are just $0$ and $-1$.
A: If the range set of the sequence contains  $k$ points each one is repeated infinitely many times (and maybe it contains other points each one reached only finite many times) then the set of limit points of the sequence is exactly those  $k$ points reached infinitely many times each.
Your example:  [1,0,-1,2,0,-2,3,0,-1,4,0,-1,... has only 2 limit points->0 and -1] is true because only 0 and -1 are reached infinitely many times, the other points are not.
EDIT. For example, the point 0 is $a_2$ and is $a_5$ and is $a_8$...
it is this $a_n$ mentioned by the definition you mentioned: there exists $a_n$ such that...
A: Just to clarify:  there certainly can be infinitely many limit points.  For example $$0,1, 0,1,2,0,1,2,3,0,1,2,3,4,....$$
In that sequence, every (non-negative) integer occurs infinitely often, hence each is a limit point.
