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Math Help - removable discontinuity and the derivative

  1. #1
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    removable discontinuity and the derivative

    I understand the proof that says a differentiable function must be continuous. I also "see" that the two sided limits of the difference quotient must be equal in order to talk about the derivative. But if the derivative is a limit why doesn't it exist about a "hole" or removable discontinuity when other limits do?
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  2. #2
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    Quote Originally Posted by zg12 View Post
    I understand the proof that says a differentiable function must be continuous. I also "see" that the two sided limits of the difference quotient must be equal in order to talk about the derivative. But if the derivative is a limit why doesn't it exist about a "hole" or removable discontinuity when other limits do?

    Because that precise limit implies the definition of the function on the point: f'(a):=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}\Longrightarrow f(a) must

    exist for the derivative to have any chance to exist.

    Tonio
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    So, if it were otherwise you would have in the numerator f(x)-undefined/(x-a) which doesn't really make since. Is this is what you're saying? I think I can buy that.
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    Quote Originally Posted by zg12 View Post
    So, if it were otherwise you would have in the numerator f(x)-undefined/(x-a) which doesn't really make since. Is this is what you're saying? I think I can buy that.

    Otherwise the definition makes no sense...not only that: the function must be defined in the point a and also in some

    open neighborhood of this point for us to be capable to begin evaluating that limit...

    Tonio
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  5. #5
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    It is possible to generalize the definition of derivative to \lim_{h\to 0}\frac{f(a+h)- f(a- h)}{2h} and avoid the question of the value at a. But that is, as I say, a generalization and not the standard "derivative".
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