Question: I'm stuck with the proof of the following:

Suppose that $(s\_n)$ converges to $s$, $(t\_n)$ converges to $t$, and $s\_n \leq t\_n \: \forall \: n$. Prove that $s \leq t$. 

I've tried starting with $s\_n \leq t\_n \: \forall : n$ and the definitions of each limit (i.e. $|s\_n - s| \leq \epsilon \: \forall \: n > N\_1$), but I'm not really getting very far. Any help is appreciated!

Answer: 
Here is a proof without contradiction: for every $\varepsilon>0$, there exists $N$ large enough so that $s-\varepsilon/2\leqslant s\_{n}\leqslant s+\varepsilon/2$ and $t-\varepsilon/2\leqslant t\_{n}\leqslant t+\varepsilon/2$ for all $n>N$.

Therefore, from the hypothesis that $s\_{n}\leqslant t\_{n}$,
$$
s-\varepsilon/2\leqslant s\_{n}\leqslant t\_{n}\leqslant t+\varepsilon/2,
$$
from which it can be concluded that 
$$
s\leqslant t+\varepsilon.
$$
Since $\varepsilon>0$ is arbitrary, the proof is done.
Since $\{s\_n\}$ converges to $s$ and $\{t\_n\}$ converges to $t$, $\{t\_n - s\_n\}$ converges to $t - s$. Since $s\_n \leq t\_n$ for all $n$, each term $t\_n - s\_n$ is nonnegative. It thus suffices to show that a sequence of nonnegative terms cannot converge to a negative limit (use proof by contradiction).