Convergent Real Sequence has Unique Limit/Proof 1

Proof
that $\sequence {s_n}$ converges to $l$ and also to $m$.

That is, suppose that:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n = l$

and:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n = m$

, assume that $l \ne m$.

Let:
 * $\epsilon = \dfrac {\size {l - m} } 2$

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

As $\sequence {s_n} \to l$:
 * $\exists N_1 \in \N: \forall n \in \N: n > N_1: \size {s_n - l} < \epsilon$

Similarly, since $\sequence {s_n} \to m$:
 * $\exists N_2 \in \N: \forall n \in \N: n > N_2: \size {s_n - m} < \epsilon$

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

We have:

This constitutes a contradiction.

It follows from Proof by Contradiction that $l = m$.