I'm learning Analysis by myself. My proof of the theorem was different to what's written in the textbook. I wanted to know if mine was also correct.
Theorem 3.2.4:
If $\mathbf{X}=(x_n)$ is a convergent sequence of real numbers and if ${x_n}\ge0$ for all $x \in \mathbb{N}$, then $x=\lim(x_n) \ge 0$
Proof:
Assume to the contrary that $x=\lim(x_n) \lt 0$. Since the sequence $(x_n)$ is convergent, $x_n$ is bounded. Thus, there exists $M \gt 0$ such that $|x_n| \le M$ for all $n \in \mathbb{N}$.
Let $\epsilon = M$. Then there exists $K$ such that $|x_n - x| < M$ for all $n \ge K$. Thus, $|x_n - x| < M \Leftrightarrow x-M\lt x_n \lt x+M$. Thus, $ x > M - x_n \ge 0 \Leftrightarrow x \ge 0$. This is contradiction.