seemingly simple limit proof

**Problem Statement:**

suppose that {$\displaystyle s_n$} and {$\displaystyle t_n$} are sequences such that $\displaystyle s_n \leq t_n$ for all $\displaystyle n$. Show that if $\displaystyle s_n \rightarrow s$ and $\displaystyle t_n \rightarrow t$, then $\displaystyle s \leq t$. Hint: an "$\displaystyle \varepsilon$-argument" works well here.

**Relevant Theorems:**

Thm 3.3-a (Baby-Rudin) - $\displaystyle \lim_{n \rightarrow \inf}(s_n+t_n)=s+t$

**My Work:**

For some $\displaystyle N_1$, when $\displaystyle n \geq N_1$ , then $\displaystyle |s-s_n| < \varepsilon$ given $\displaystyle \varepsilon > 0$.

For some $\displaystyle N_2$, when $\displaystyle n \geq N_2$ , then $\displaystyle |t- t_n| < \varepsilon$ given $\displaystyle \varepsilon > 0$.

now, $\displaystyle \lim_{n \rightarrow \inf}s_n=s$ and $\displaystyle \lim_{n \rightarrow \inf}t_n=t$.

From thm 3.3, we have that $\displaystyle \lim_{n \rightarrow \inf}(t_n-s_n)=t-s$ and since $\displaystyle (t_n-s_n) \geq 0$ for all $\displaystyle n$, then $\displaystyle t-s \geq 0$