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).
Let's consider $x\_n=t\_n-s\_n$, so your task can be formulate as following: if for converged sequence $x\_n$ holds $\forall n \in \mathbb{N}, x\_n \geqslant 0$, then $\lim\limits\_{n \to \infty}x\_n=x\geqslant 0$.

Let's assume, that $x<0$. Then also $\frac{x}{2}<0$. Now if we take $\varepsilon=-\frac{x}{2}>0$, then from convergence we have $\exists N \in \mathbb{N}$, such that $\forall n>N$ we have $|x\_n-x|<\varepsilon \Leftrightarrow x-\varepsilon<x\_n<x+\varepsilon=x-\frac{x}{2}=\frac{x}{2}<0$ and we come to contradiction.