Convergent Real Sequence is Bounded/Proof 2

Theorem
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

Let $l \in A$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.

Then $\left \langle {x_n} \right \rangle$ is bounded.

That is, all convergent real sequences are bounded.

Proof
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

Let $x_n \to l$ as $n \to \infty$.

To show that $\left \langle {x_n} \right \rangle$ is bounded sequence, we need to find $K$ such that:
 * $\forall n \in \N: \left|{x_n}\right| \le K$

Since $\left \langle {x_n} \right \rangle$ converges:
 * $\forall \epsilon > 0: \exists N: n > N \implies \left|{x_n - l}\right| < \epsilon$

In particular, this is true when $\epsilon = 1$.

That is:
 * $\exists N_1: \forall n > N_1: \left|{x_n - l}\right| < 1$

By Backwards Form of Triangle Inequality:
 * $\forall n > N_1: \left|{x_n}\right| - \left|{l}\right| \le \left|{x_n - l}\right| < 1$

That is:
 * $\left|{x_n}\right| < \left|{l}\right| + 1$

So we set:
 * $K = \max \left\{ {\left| {x_1} \right|, \left| {x_2} \right|, \ldots, \left| {x_{N_1}} \right|, \left| {l} \right| + 1} \right\}$

and the result follows.