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!
 A: 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.
A: 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).
A: 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.
A: 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?)
A: 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.
