Proof that limit points are unique I ams having problems to prove that the limits points of a sequence are unique. For example given the following sequence
\begin{equation}
x_n=(-1)^n+\frac{1}{n}
\end{equation}
To find the limit points, I establish these subsequences
\begin{equation}
y_n=x_{2n}=(-1)^{2n}+\frac{1}{2n}=1+\frac{1}{2n}
\end{equation}
\begin{equation}
z_n=x_{2n+1}=(-1)^{2n+1}+\frac{1}{2n+1}=-1+\frac{1}{2n+1}
\end{equation}
Where it can be seen that
\begin{equation}
y_n\rightarrow 1
\end{equation}
\begin{equation}
z_n\rightarrow -1
\end{equation}
My question is how can I prove that these are the only two limit points of the sequence.
I suppose that it is related to the fact that
\begin{equation}
Image(f(n))+Image(g(n))=\mathbb{N}
\end{equation}
Being
\begin{equation}
f(n)=2n\; \; \; \;  
g(n)=2n+1
\end{equation}

Suppose that $L$ is another number, not equal to $1$ or to $-1$. Then we can find neighborhoods of $1$, $-1$, and $L$, all mutually disjoint. Since the sequence is eventually inside the union of the neighborhoods of $1$ and $-1$, then it's eventually outside the neighborhood of $L$, so $L$ is not a limit point.
