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: 
Hint: Suppose $s>t$ for a contradiction. Then intuitively, for $n$ large, $s\_n$ is very close to $s$ and $t\_n$ is very close to $t$, so $s\_n$ would have to be greater than $t\_n$. Can you find an $\epsilon$ so that if you knew $s\_n$ were within $\epsilon$ of $s$ and $t\_n$ were within $\epsilon$ of $t$, then $s\_n$ would be greater than $t\_n$? (If you have trouble doing this, you might consider a concrete example: suppose $s=1$ and $t=0$. Then what does $\epsilon$ need to be?)
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.