# Limit Proof

by dogma
Tags: limit, proof
 P: 35 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
 P: 83 $$\lim_{x\rightarrow x_o} f(x) = - A$$ 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 $$\epsilon = A$$?

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