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}

Take $a\in\Bbb R\setminus\{1,-1\}$. Then, take $\varepsilon=\min\{|a+1|,|a-1|\}$. Since $a\ne\pm1$, $\varepsilon>0$. Since $\lim_{n\to\infty}x_{2n}=1$ and $\lim_{n\to\infty}x_{2n-1}=-1$, there is some $N_e\in\Bbb N$ such that$$n\geqslant N_e\implies|x_{2n}-1|<\frac\varepsilon2$$and there is some $N_o\in\Bbb N$ such that$$n\geqslant N_o\implies|x_{2n-1}+1|<\frac\varepsilon2.$$But then, since each natural number is either even or odd, if $n$ is large enough, then you have $|x_n+1|,|x_n-1|<\frac\varepsilon2$. So, since $\varepsilon\leqslant|a+1|,|a-1|$, $|x_n-a|>\frac\varepsilon2$. Since the inequality $|x_n-a|>\frac\varepsilon2$ takes place for every $n$ is large enough, $a$ cannot be the limit of a subsequence of $(x_n)_{n\in\Bbb N}$.
