a real convergent sequence has a unique limit point How to show that a real convergent sequence has a unique limit point viz. the limit of the sequene?
I've used the result several times but I don't know how to prove it!
Please help me!

I am assuming that limit points are defined as in Section $6.4$ of the book Analysis $1$ by the author Terence Tao. We assume that the sequence of real numbers $(a_{n})_{n=m}^{\infty}$ converges to the real number $c$. Then we have to show that $c$ is the unique limit point of the sequence. First, we shall show that $c$ is indeed a limit point. By the definition of convergence we have
$$\forall\epsilon > 0\:\exists N_{\epsilon}\geq m\:\text{s.t.}\:\forall n\geq N_{\epsilon}\:|a_{n}-c|\leq\epsilon$$
We need to prove that $\forall\epsilon >0\:\forall N\geq m\:\exists n\geq N\:\text{s.t.}\:|a_{n}-c|\leq\epsilon$. Fix $\epsilon$ and $N$. If $N\leq N_{\epsilon}$ then we could pick and $n\geq N_{\epsilon}$. If $N>N_{\epsilon}$ then picking any $n\geq N > N_{\epsilon}$ would suffice.
Now, we shall prove that $c$ is the unique limit point. Let us assume for the sake of contradiction that $\exists$ a limit point $c'\in\mathbb{R}$ and $c'\neq c$. Since the sequence converges to $c$, $\forall\epsilon > 0\:\exists M_{\epsilon}\geq m$ such that $\forall n\geq M_{\epsilon}$ we have $|a_{n}-c|\leq \epsilon/2$. Further, since $c'$ is a limit point, $\forall\epsilon > 0$ and $\forall N\geq m\:\exists k_{N}\geq N$ such that $|a_{k_{N}}-c'|\leq \epsilon/2$. In particular, $|a_{k_{M_{\epsilon}}}-c'|\leq \epsilon/2$ and $|a_{k_{M_{\epsilon}}}-c|\leq \epsilon/2$. Using the triangle inequality we get $\forall\epsilon > 0\: |c-c'|\leq\epsilon\implies |c-c'| = 0\implies c = c'$ which contradicts the assumption.
