Suppose $(a_n),(b_n)$ are two convergent sequences such that there exists some $N\in\mathbb{N}$ such that $a_n\geq b_n$ for all $n\geq N$. Then $$\lim_{n \to \infty} a_n \geq \lim_{n \to \infty} b_n.$$
This is an obvious fact but I'm not sure how to formalize the proof.
Take the subsequences $a_{n_k}$ and $b_{n_k}$ which are defined to be the $N$th tail of $(a_n)$ and $(b_n)$ respectively. Since every subsequence of a convergent sequence converges to the same limit, we have $$\lim a_n=\lim a_{n_k}\geq \lim b_{n_k}=\lim b_n.$$
Is there a better proof than this?
Suppose $a_n \to a$ and $b_n \to b$ and suppose $a<b$. Fix $\varepsilon:=\frac{b-a}{3}$. Then $$ a_n<a+\varepsilon<b-\varepsilon<b_n $$ for all $n$ sufficiently large. This contradicts the hypothesis $a_n \ge b_n$ eventually.
Ps. I wouldn't call it "better".