If $(x_{n})$ as sequence with $x_{n} \leq b$ for all $n$, then $\text{lim}_{n \to \infty} x_{n} \leq b $ if limit exists.
I see this fact being used very often and it seems fairly trivial but I am wondering how it can be proven rigorously.
Proof attempt:
Suppose that the limit exists.
Suppose $x_{n}\leq b$ for all $n \in \mathbb{N}$. Then by definition of limit, for any arbitrary $\epsilon>0$, we can find $N \in \mathbb{N}$ such that for every $n^*>N$ we have $|(\text{lim}_{n \to \infty} x_{n})-x_{n^*}|< \epsilon$. Then $\text{lim}_{n \to \infty} x_{n}<\epsilon +x_{n^*} \leq \epsilon +b$
So $\text{lim}_{n \to \infty} x_{n}< \epsilon + b$ where $\epsilon$ is arbitrary.
This implies that at most, $\text{lim}_{n \to \infty} x_{n} \leq b$