epsilon delta proof by contradiction

[chroye]

I need to use the definition of continuity and limits to show:

let f : R → R be a function continuous in x = 0, with the property that if x!=0, f(x) ≥ 0. This makes f(0) ≥ 0. This part is what i need to prove I assume.
(Only supposing f is continuous at point a, not necesserily around a).

Basically I have extracted from the text: since f is cont. in x= 0, we have: lim as x approaches 0 of f(x) = f(0). This gives me: |f(x)-f(0)| < E
Furthermore i figure the easiest option is to do a proof by contradiction using epsilon delta, and even tho i know how to use epsilon-delta, i cant really figure out how to arrive at a contradiction/ how to write this up. Do i just limit delta to some arbitrary value? When i tried for delta <=1 I didn't seem to get a contradiction, at any rate i got really confused. Any help would be appreciated, thanks!

 

[Dr.Peterson]

Let's start by being clearer about how you're starting.

What are you assuming for the sake of contradiction?

 

[Steven G]

It seems that you want to prove that if f(x) > 0, the f(x)>0. I doubt that you want to show this as it is trivial.

 

[pka]

let f : R → R be a function continuous in x = 0, with the property that if x!=0, f(x) ≥ 0. This makes f(0) ≥ 0. This part is what i need to prove I assume. (Only supposing f is continuous at point a, not necesserily around a).

If we can show that \(f(0)<0\) leads to a conduction then we know that \(f(x)\ge 0\)
Suppose that \(f(0)<0\) Now I cannot give you a detailed answer. I have just been warned by a know-nothing moderator (I have not heard yet from Ted). So here goes: let \(\delta=f(0)<0\). Clearly by continuity \(\exists t\in (-\delta,\delta)\) such that \(f(t)<0\) but that is a contradiction.

 

[chroye]

If we can show that \(f(0)<0\) leads to a conduction then we know that \(f(x)\ge 0\)
Suppose that \(f(0)<0\) Now I cannot give you a detailed answer. I have just been warned by a know-nothing moderator (I have not heard yet from Ted). So here goes: let \(\delta=f(0)<0\). Clearly by continuity \(\exists t\in (-\delta,\delta)\) such that \(f(t)<0\) but that is a contradiction.

Thanks for the reply! What have you been warned about? I think i get what you're getting at here, i'll sit down and think about it more thoroughly when i get time. Not immediately obvious to me why you set delta equal to f(0) tho, could you explain that?
In a formal proof doesn't delta have to be expressed by Epsilon or a min { } value? Also delta has to be greater than 0 right?