Limits of Inequality: Proving by Contradiction

[Buri]

Homework Statement

If f(x) <= g(x) then lim[x->a] f(x) <= lim[x->a] g(x) provided that both of these limits exist.


2. The attempt at a solution

I've been able to prove it by contradiction. So I assumed that l = lim[x->a] f(x) > lim[x->a] g(x) = m. Therefore, l - m > 0 and I could choose epsilon = (l - m)/2 and the contradiction follows. However, what I need someone to help me with is how can I "see" that (l - m)/2 will actually work (i.e. yield a contradiction)? I just immediately thought of using this choice of epsilon and a contradiction followed. But going back to the problem now, I have no clue why I choose this epsilon. Can someone be able to help me as to why this one actually works?

Thanks

 

[Mark44]

(l - m)/2 is half the difference of l and m.

 

[Buri]

(l - m)/2 is half the difference of l and m.

LOL! Obviously...