Limit Proof


by dogma
Tags: limit, proof
dogma
dogma is offline
#1
Sep28-05, 12:43 PM
P: 35
Hi all.

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

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

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

Assume [tex]\lim_{x\rightarrow x_o} f(x) = A < 0[/tex]

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

Can someone please point me on the right direction?

Thanks,
dogma
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Timbuqtu
Timbuqtu is offline
#2
Sep28-05, 01:59 PM
P: 83
[tex]\lim_{x\rightarrow x_o} f(x) = - A[/tex] means: for all [tex]\epsilon > 0[/tex] there exist a [tex]\delta > 0[/tex] such that [tex]f( \left]x_0 - \delta, x_0 + \delta \right[) \subset \left]-A-\epsilon, -A+\epsilon \right[ [/tex]. So in particular what do we get for [tex]\epsilon = A[/tex]?


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