# Thread: seemingly simple limit proof

1. ## seemingly simple limit proof

Problem Statement:
suppose that {$\displaystyle s_n$} and {$\displaystyle t_n$} are sequences such that $\displaystyle s_n \leq t_n$ for all $\displaystyle n$. Show that if $\displaystyle s_n \rightarrow s$ and $\displaystyle t_n \rightarrow t$, then $\displaystyle s \leq t$. Hint: an "$\displaystyle \varepsilon$-argument" works well here.

Relevant Theorems:
Thm 3.3-a (Baby-Rudin) - $\displaystyle \lim_{n \rightarrow \inf}(s_n+t_n)=s+t$
My Work:
For some $\displaystyle N_1$, when $\displaystyle n \geq N_1$ , then $\displaystyle |s-s_n| < \varepsilon$ given $\displaystyle \varepsilon > 0$.
For some $\displaystyle N_2$, when $\displaystyle n \geq N_2$ , then $\displaystyle |t- t_n| < \varepsilon$ given $\displaystyle \varepsilon > 0$.
now, $\displaystyle \lim_{n \rightarrow \inf}s_n=s$ and $\displaystyle \lim_{n \rightarrow \inf}t_n=t$.
From thm 3.3, we have that $\displaystyle \lim_{n \rightarrow \inf}(t_n-s_n)=t-s$ and since $\displaystyle (t_n-s_n) \geq 0$ for all $\displaystyle n$, then $\displaystyle t-s \geq 0$

2. Originally Posted by DontKnoMaff
Problem Statement:
suppose that {$\displaystyle s_n$} and {$\displaystyle t_n$} are sequences such that $\displaystyle s_n \leq t_n$ for all $\displaystyle n$. Show that if $\displaystyle s_n \rightarrow s$ and $\displaystyle t_n \rightarrow t$, then $\displaystyle s \leq t$. Hint: an "$\displaystyle \varepsilon$-argument" works well here.

Relevant Theorems:
Thm 3.3-a (Baby-Rudin) - $\displaystyle \lim_{n \rightarrow \inf}(s_n+t_n)=s+t$
My Work:
For some $\displaystyle N_1$, when $\displaystyle n \geq N_1$ , then $\displaystyle |s-s_n| < \varepsilon$ given $\displaystyle \varepsilon > 0$.
For some $\displaystyle N_2$, when $\displaystyle n \geq N_2$ , then $\displaystyle |t- t_n| < \varepsilon$ given $\displaystyle \varepsilon > 0$.
now, $\displaystyle \lim_{n \rightarrow \inf}s_n=s$ and $\displaystyle \lim_{n \rightarrow \inf}t_n=t$.
(*) From thm 3.3, we have that $\displaystyle \lim_{n \rightarrow \inf}(t_n-s_n)=t-s$ and since $\displaystyle (t_n-s_n) \geq 0$ for all $\displaystyle n$, then $\displaystyle t-s \geq 0$
How does (*) follow from any of the above? What you're saying is true (that if a convergent sequence is always non-negative, then its limit is non-negative), but it's equivalent to what you're trying to show. And I don't see how it follows from what you've written. (What have you done with $\displaystyle \epsilon, N_1, \mbox{and }N_2$, apart from summoning them into existence?)

A proof by contradiction works well. Suppose that $\displaystyle a_n \geq 0 \ \forall n$ and that $\displaystyle a_n \to a$. Suppose that $\displaystyle a<0$. Take $\displaystyle \epsilon>0$ small enough that $\displaystyle a+\epsilon < 0$. What do you get if you find $\displaystyle N$ large enough that $\displaystyle |a_n-a|<\epsilon$ for $\displaystyle n\geq N$? (Break up the absolute value into two inequalities; one of them will contradict the non-negativity of $\displaystyle a_n$).

Then take $\displaystyle a_n$ to be the sequence $\displaystyle t_n-s_n$, as you did.

3. thanks alot. I am horrible at writing these proofs. OK here is what I've got now:

take {$\displaystyle a_n$} where $\displaystyle a_n=t_n-s_n$ for all $\displaystyle n$ and {$\displaystyle a_n$} $\displaystyle \rightarrow a=t-s$.
We have assumed that $\displaystyle s_n \leq t_n$ which yields $\displaystyle t_n-s_n \geq 0$, so $\displaystyle a_n \geq 0$ for all $\displaystyle n$.
We now want $\displaystyle a>0$.
Suppose $\displaystyle a<0$, take $\displaystyle \epsilon > 0$ s.t. $\displaystyle a+\epsilon < 0$.
Now take $\displaystyle N$ s.t. when $\displaystyle n \geq N$, then , $\displaystyle |a_n-a| < \epsilon$.
Now since $\displaystyle |a| > \epsilon$ and $\displaystyle a <0$ and $\displaystyle a_n >0$, this is a contradiction and so $\displaystyle a \geq 0$ which gives $\displaystyle t-s \geq 0$

I think that works now.

4. Everything is fine until the last line.
What you should have written is that since $\displaystyle |a_n-a| < \epsilon$, we have $\displaystyle -\epsilon < a_n-a < \epsilon$ (this is by definition of the absolute value). But $\displaystyle a_n-a < \epsilon \Rightarrow a_n < a+\epsilon < 0$, which contradicts $\displaystyle a_n \geq 0$.