Continuous function / epsilon-delta

[zeebo17]

Homework Statement

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.

Homework Equations

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

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!

 

[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.

 

[zeebo17]

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

 

[HallsofIvy]

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

 

[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&gt;0 such that |x-a|&lt; \delta implies that |h(x)-h(a)|&lt; \epsilon. In this case h(a)-h(x) > 0, so 0 &lt; h(a)-h(x) &lt; \epsilon.

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