- #1
Buri
- 273
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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