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: 
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.
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.