# Homework Help: Another epsilon-delta proof.

1. Oct 18, 2007

### bjgawp

1. The problem statement, all variables and given/known data
I've got another question involving epsilon-delta proofs, one that is less concrete:
Prove that if g(x) $$\geq$$ 0 near c and $$\lim_{x \to c} g(x) = M$$ then M $$\geq$$ 0. Furthermore, if g(x) > 0 does it follow that M > 0?

2. Relevant equations

3. The attempt at a solution
Starting off with some preliminary work:
Let $$\epsilon$$ > 0. We must find $$\delta$$ > 0 such that |g(x) - M| < $$\epsilon$$ whenever 0 < |x - c| < $$\delta$$

Does anyone have a hint they could provide? I'm not even sure how the end result of this proof is suppose to look like so that's a major set-back in proving this. Simpler, concrete examples require finding a $$\delta$$ in terms of $$\epsilon$$ but I don't know how that would apply here.

2. Oct 18, 2007

### ZioX

I would do a proof by contradiction.

The furthermore part is true since g is strictly positive on a neighbourhood around c. Although, what can you say about the case where $\lim_{x\to+\infty}g(x)=M$ when g(x)>0 for all x. Is M>0? If you don't know how to evaluate these kinds of limits just think of a proof/disproof informally.

Last edited: Oct 18, 2007
3. Oct 18, 2007

### bjgawp

Hmm following your advice, the most I can come up with is trying to analyze the following
M - $$\epsilon$$ < g(x) < M + $$\epsilon$$
which isn't a lot as the epsilon poses a problem in coming to a contradiction (such as M > g(x) or M > 0). Anything I'm missing ?

Last edited: Oct 18, 2007
4. Oct 19, 2007

### ZioX

Suppose it was negative. Then we know by our delta/epsilon definition of convergence that g(x) must be negative as x gets gets close to c.

Prove this formally.

5. Oct 19, 2007

### bjgawp

Hmm, sorry for my lack of knowledge but I haven't gone through anything involving convergence (Cauchy sequences?). We've just done a few lectures about epsilon-delta proofs but it's hard to apply what we learn from simple,concrete examples to the general cases. In any case, with the suggestion of doing a proof of contradiction, this is what I've gotten:
$$|g(x)-M|<\epsilon$$
$$g(x)+|M|<\epsilon$$ (since L < 0 and f(x) $$\geq$$ 0)
$$0<|M|<\epsilon-g(x)$$
$$g(x)-\epsilon<M<0<\epsilon-g(x)$$

At this point, I can see that e - g(x) can certainly be less than zero if we restrict epsilon small enough. Does this seem right? I know I need to solidify this "proof", if you can call it,that even more.

Last edited: Oct 19, 2007
6. Oct 19, 2007

### Dick

Assume M<0. How about choosing epsilon=|M/2|? if |g(x)-M|<|M/2| then since we've assumed M negative, this forces g(x) to be negative.

7. Oct 19, 2007

### ZioX

I made a mistake, the furthermore part is not true. I was picturing continuous functions in my head for some reason. Let that be a hint.