# Two simple limit question

Q.1 Assume $s_n\ge a$ for all but finitely many n, then $\lim s_n\ge a$.

I think using contradiction can prove it but i wonder if there is a more direct proof.

Q.2 suppose that there exists $N_0$ such that $s_n\le t_n$for all $n>N_0$, prove that if $\lim s_n$ and $\lim t_n$ exist, then $\lim s_n\le \lim t_n$.

-
What is contraction? –  timur Jul 27 '12 at 4:22
Q.1 follows immediately from the definition of a limit, assuming that $s_n$ has a limit, of course. Q.2 follows immediately from Q.1 by looking at the sequence $t_n-s_n$ and the constant $a=0$. –  copper.hat Jul 27 '12 at 4:26
In the italian literature, we call these facts "Teoremi della permanenza del segno". The proofs are exactly those you suggest. –  Siminore Aug 26 '12 at 8:37

Q.2 follows from Q.1, using $a=0$ and the sequence of general term $t_n-s_n$.
For a proof of Q.1 not by contradiction, define $s=\lim\limits_{n\to\infty}s_n$ and consider any $b\lt a$. Then $\varepsilon=a-b$ is such that $\varepsilon\gt0$. Since $s_n\geqslant a$ for every $n$ large enough, $|s_n-b|\geqslant s_n-b\geqslant\varepsilon$ for every $n$ large enough, hence $s\ne b$. Since $s\ne b$ for every $b\lt a$, $s\geqslant a$. (Note that this assumes at the onset that the limit $s$ exists, otherwise the conclusion must be downplayed to $\liminf\limits_{n\to\infty}s_n\geqslant a$.)
Actually, i got a question, what do finitely many $n$ mean? what about infinitely many $n$, why the condition is needed?? –  Mathematics Sep 25 '12 at 13:42
The assertion that $s_n\geqslant a$ for all but finitely many $n$ means that the set $\{n\mid s_n\lt a\}$ is (empty or) finite. –  Did Sep 25 '12 at 13:53
For Q.1, let $N = \max\{k : s_k < a\}$. Then $n \ge N$ imples $s_n \ge a$.