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A general e-d proof

  1. Oct 12, 2003 #1
    What I tried to do with this proof is show that if a function is differentiable at a point, then the limit of the function at this point is the value of the function at that point. This seems like a pretty obvious point (since differentiability depends on continuity and continuity depends on the existence of the limit), so I tried to go a step further and identify a specific delta for each epsilon that would guarantee the existence of this limit. I was hopefully successful.
    I spent about two hours working out the details (don't worry - it's not too long) and when I finally finished I was too exhausted to see if my proof stood up to a bunch of specific tests. Anyway, I was hoping someone more experience in math could check my proof. If it's wrong, show me where.
    The purpose of this proof is to make finding deltas easy when I know that a function is differentiable. This is acceptable because finding delta is not a necessary part of a limit proof. If my first guess for delta work for all epsilons, then I have completed the proof.

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  3. Oct 13, 2003 #2


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    Have you had the "mean value theorem" in calculus yet?

    That looks to me like what you are trying to do.

    The mean value theorem says that if f is continuous on some interval [a,b]and differentiable on (a,b) then there exist some point c in (a,b) such that (f(b)- f(a))/(b-a)= f'(c). From that,
    f(b)- f(a)= f'(c)(b-a). In particular, if f' is continuous on (a,b) (which will always be true if f is also differentiable at a and b), then |f'(c)| has a maximum value on [a,b], say M, and we can say
    |f(b)- f(a)|<= M|b-a|. Finally, given epsilon> 0, we can take
    delta= epsilon/M.
  4. Oct 13, 2003 #3
    Yes, I have studied the mean value theorem before (although not in the form you presented), but I was not thinking of this proof in that context. Does the mean value theorem allow you to find the value of M?
  5. Oct 19, 2003 #4


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    NO, there is no simple formula for M but it doesn't matter- it's a constant and so as x-> a M|x-a| goes to 0.
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