Convergent Real Sequence has Unique Limit/Proof 1

Theorem
Let $\left \langle {s_n} \right \rangle$ be a real sequence.

Then $\left \langle {s_n} \right \rangle$ can have at most one limit.

Proof
Suppose that $\left \langle {s_n} \right \rangle$ converges to $l$ and also to $m$.

That is, suppose $\displaystyle \lim_{n \to \infty} x_n = l$ and $\displaystyle \lim_{n \to \infty} x_n = m$.

Assume that $l \ne m$, and let:
 * $\epsilon = \dfrac {\left\vert{l - m}\right\vert} 2$

As $l \ne m$, it follows that $\epsilon > 0$.

Therefore, since $\left \langle {s_n} \right \rangle \to l$:
 * $\exists N_1 \in \N: \forall n \in \N: n > N_1: \left\vert{s_n - l}\right\vert < \epsilon$

Similarly, since $\left \langle {s_n} \right \rangle \to m$:
 * $\exists N_2 \in \N: \forall n \in \N: n > N_2: \left\vert{s_n - m}\right\vert < \epsilon$

Now set $N = \max\left\{{N_1, N_2}\right\}$.

We have:

This constitutes a contradiction.

Therefore, it must be that $l = m$.