Proving If f(x)>0 for All x for Dogma

[dogma]

Hi all.

I'm doing some self studying on limits, and...I have the following problem with this problem...

Prove: If f(x)>0 for all x, then \lim_{x\rightarrow x_o} f(x)\geq 0 for any x_o

I'm assuming the best way to prove this is through contradiction:

Assume \lim_{x\rightarrow x_o} f(x) = A < 0

This as far I get before vapor lock sets in. I guess I need to find an appropriate \epsilon and then try to show/not show that f(x) < 0 for at least one x.

Can someone please point me on the right direction?

Thanks,
dogma

 

[Timbuqtu]

\lim_{x\rightarrow x_o} f(x) = - A means: for all \epsilon > 0 there exist a \delta > 0 such that f( \left]x_0 - \delta, x_0 + \delta \right[) \subset \left]-A-\epsilon, -A+\epsilon \right[. So in particular what do we get for \epsilon = A?