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!

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.
