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'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!

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?)
