Convergent Real Sequence has Unique Limit

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 $$\lim_{n \to \infty} x_n = l$$ and $$\lim_{n \to \infty} x_n = m$$.

Let:
 * $$\epsilon = \frac {\left|{l - m}\right|} 2$$

Since $$\left \langle {s_n} \right \rangle \to l$$:
 * $$\exists N_1 \in \N: \forall n \in \N: n > N_1: \left|{s_n - l}\right| < \epsilon$$

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

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

We have:

$$ $$ $$ $$

From this contradiction it follows that $$l = m$$.

Alternative Proof
From the fact that the real number line is a metric space, we can directly use Sequence in Metric Space has One Limit at Most‎.