# For $s_n \leq t_n$ above specific n, prove $\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$

The question: Suppose there exists $$N_0$$ such that $$s_n \leq t_n$$ for all $$n > N_0$$. Prove that if $$\lim_{n\to\infty} s_n$$ and $$\lim_{n\to\infty} t_n$$ exists, then $$\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$$.

My attempt at a proof:

Let $$\lim_{n\to\infty} s_n = s$$ and $$\lim_{n\to\infty} t_n = t$$. Assume, for purposes of contradiction, that $$s > t$$. Since $$\lim_{n\to\infty} s_n$$ exists, it must be true that there exists $$N_1$$ for all $$\epsilon > 0$$ such that $$n > N_1$$ implies $$\mid s_n - s \mid < \epsilon$$. Since, by premise, $$s - t > 0$$, we may choose $$\epsilon = s - t$$ and there must exist $$N_1$$ satisfying $$\mid s_n - s \mid < \epsilon$$, i.e. $$\mid s_n - s \mid < s - t$$, and thus by the absolute value properties, $$s - (s-t) < s_n < s + (s-t)$$, which directly implies $$t < s_n$$ for sufficiently large $$n$$. Now we examine the implications on $$\lim_{n\to\infty} t_n$$ exists, we may select any $$\epsilon > 0$$ to satisfy $$\mid t_n - t \mid < s_n - t$$. Since $$s_n - t > 0$$ for sufficiently large $$n$$, we may set $$\epsilon = s_n - t$$ for some $$s_n$$, however this would imply some $$N$$ exists such that $$n > N$$ implies $$\mid t_n - t \mid < s_n - t$$, i.e. $$t_n < s_n - t + t$$, that is, $$t_n < s_n$$ for all sufficiently large $$n$$, which is a contradiction. Thus $$\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$$.

I'm certain I got lost in the details somewhere. Could someone point out errors that I've made?

Your idea of the proof is correct, though you made it a bit too complicated with the technical part. Here is a clear way to write it. Assume $$s>t$$. Then there exists $$\epsilon>0$$ such that $$s-\epsilon>t+\epsilon$$. Now you know there is $$N_1\in\mathbb{N}$$ such that $$s_n>s-\epsilon$$ for all $$n\geq N_1$$. Also there is $$N_2\in\mathbb{N}$$ such that $$t_n for all $$n\geq N_2$$. Now let $$N=max\{N_0,N_1,N_2\}$$. For all $$n\geq N$$ you have:
$$s_n>s-\epsilon>t+\epsilon>t_n$$