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Another epsilon-delta proof.

  1. Oct 18, 2007 #1
    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) [tex]\geq[/tex] 0 near c and [tex]\lim_{x \to c} g(x) = M [/tex] then M [tex]\geq[/tex] 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 [tex]\epsilon[/tex] > 0. We must find [tex]\delta[/tex] > 0 such that |g(x) - M| < [tex]\epsilon[/tex] whenever 0 < |x - c| < [tex]\delta[/tex]

    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 [tex]\delta[/tex] in terms of [tex]\epsilon[/tex] but I don't know how that would apply here.
     
  2. jcsd
  3. Oct 18, 2007 #2
    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 [itex]\lim_{x\to+\infty}g(x)=M[/itex] 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
  4. Oct 18, 2007 #3
    Hmm following your advice, the most I can come up with is trying to analyze the following
    M - [tex]\epsilon[/tex] < g(x) < M + [tex]\epsilon[/tex]
    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 o_O?
     
    Last edited: Oct 18, 2007
  5. Oct 19, 2007 #4
    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.
     
  6. Oct 19, 2007 #5
    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:
    [tex]|g(x)-M|<\epsilon[/tex]
    [tex]g(x)+|M|<\epsilon[/tex] (since L < 0 and f(x) [tex]\geq[/tex] 0)
    [tex]0<|M|<\epsilon-g(x)[/tex]
    [tex]g(x)-\epsilon<M<0<\epsilon-g(x)[/tex]

    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
  7. Oct 19, 2007 #6

    Dick

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    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.
     
  8. Oct 19, 2007 #7
    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.
     
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