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Continuous function / epsilon-delta

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

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


    2. Relevant equations

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

    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. jcsd
  3. Nov 16, 2009 #2
    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.
     
  4. Nov 16, 2009 #3
    Yes, that makes sense intuitively. I'm still unsure how to start writing a formal proof though.
     
  5. Nov 16, 2009 #4

    HallsofIvy

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    In the definition of "limit", let [math]\epsilon= \frac{h(a)}{2}[/itex].
     
  6. Nov 16, 2009 #5
    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 [tex] \delta>0 [/tex] such that [tex] |x-a|< \delta [/tex] implies that [tex] |h(x)-h(a)|< \epsilon [/tex]. In this case h(a)-h(x) > 0, so [tex] 0 < h(a)-h(x) < \epsilon [/tex].

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