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: 
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).
Hint: Suppose $s>t$ for a contradiction. Then intuitively, for $n$ large, $s\_n$ is very close to $s$ and $t\_n$ is very close to $t$, so $s\_n$ would have to be greater than $t\_n$. Can you find an $\epsilon$ so that if you knew $s\_n$ were within $\epsilon$ of $s$ and $t\_n$ were within $\epsilon$ of $t$, then $s\_n$ would be greater than $t\_n$? (If you have trouble doing this, you might consider a concrete example: suppose $s=1$ and $t=0$. Then what does $\epsilon$ need to be?)