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: 
Here is a proof without contradiction: for every $\varepsilon>0$, there exists $N$ large enough so that $s-\varepsilon/2\leqslant s\_{n}\leqslant s+\varepsilon/2$ and $t-\varepsilon/2\leqslant t\_{n}\leqslant t+\varepsilon/2$ for all $n>N$.

Therefore, from the hypothesis that $s\_{n}\leqslant t\_{n}$,
$$
s-\varepsilon/2\leqslant s\_{n}\leqslant t\_{n}\leqslant t+\varepsilon/2,
$$
from which it can be concluded that 
$$
s\leqslant t+\varepsilon.
$$
Since $\varepsilon>0$ is arbitrary, the proof is done.
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.