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: 
Suppose $s>t$ write $c=s-t$ there exists $N$ such that $n>N$ implies $|s\_n-s|<c/4$. There exists $N'$ such that $n>N'$ implies $|t\_n-t|<c/4$, take $n>\sup(N,N')$ $s\_n-t\_n=s\_n-s+s-t+t-t\_n\geq s-t-|s\_n-s|-|t-t\_n|\geq c-c/4-c/4\geq c/2$ contradiction.
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).