# Continuous function / epsilon-delta

1. Nov 16, 2009

### zeebo17

1. The problem statement, all variables and given/known data

Let $$h: \Re \rightarrow \Re$$ be a continuous function such that h(a)>0 for some $$a \in \Re$$. Prove that there exists a $$\delta >0$$ such that h(x)>0 provided that $$|x-a|< \delta$$.

2. Relevant equations

Continuity of h means that there exists and $$\epsilon >0$$ such that $$|h(x)-h(a)| < \epsilon$$ provided that $$|x-a| < \delta$$

3. The attempt at a solution

I tried starting with the definition of continuity and perhaps the intermediate value theorem, but I haven't been able to get started.

Any suggestions on how to get started?
Thanks!

2. Nov 16, 2009

### cap.r

so because of the continuity of your function you know that there exists a delta greater than zero that will satisfy this. h(x) is greater than zero because you know that h(a) is greater than zero and if x is close to a the function has to stay positive. if it jumped around then it wouldn't be continuous.

3. Nov 16, 2009

### zeebo17

Yes, that makes sense intuitively. I'm still unsure how to start writing a formal proof though.

4. Nov 16, 2009

### HallsofIvy

In the definition of "limit", let [math]\epsilon= \frac{h(a)}{2}[/itex].

5. Nov 16, 2009

### zeebo17

Ok, Would this work:

If h(x)> h(a) then h(x)>0 since h(x)>h(a)>0.

If h(x)<h(a) then h(a)-h(x) > 0. Then because h is a continuous function there exists a $$\delta>0$$ such that $$|x-a|< \delta$$ implies that $$|h(x)-h(a)|< \epsilon$$. In this case h(a)-h(x) > 0, so $$0 < h(a)-h(x) < \epsilon$$.

If $$\epsilon = \frac{h(a)}{2}$$ then
$$0< h(a)-h(x)< \epsilon$$
$$0< h(a)-h(x)< \frac{h(a)}{2}$$
$$0< \frac{h(a)}{2}< h(x)$$
and therefore h(x) is greater then zero.