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!

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).
