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Math Help - A small part of a chain rule proof I don't get

  1. #1
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    A small part of a chain rule proof I don't get

    Hello,

    I've been trying to get my head round part of a proof and I don't see the following implication.

    Suppose we have

    \lim_{x \to x_0} E(g(x)) = E(g(x_0)) = 0

    then does it follow that

    \lim_{x \to x_0}[f'(g(x_0) + E(g(x)))]\frac{g(x) - g(x_0)}{x - x_0} = f'(g(x_0))g'(x_0)?

    Since doesn't the above require that f' is continuous at g(x_0), and if it does, why is this necessarily true?

    Thanks

    Stonehambey
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Stonehambey View Post
    Hello,

    I've been trying to get my head round part of a proof and I don't see the following implication.

    Suppose we have

    \lim_{x \to x_0} E(g(x)) = E(g(x_0)) = 0

    then does it follow that

    \lim_{x \to x_0}[f'(g(x_0) + E(g(x)))]\frac{g(x) - g(x_0)}{x - x_0} = f'(g(x_0))g'(x_0)?

    Since doesn't the above require that f' is continuous at g(x_0), and if it does, why is this necessarily true?

    Thanks

    Stonehambey
    There's too little context here, but from what I can tell you used the continuity of f at g(x_0) to conclude that \displaystyle \lim_{x\to x_0}f(g(x))=f(g(x_0)) it seems. Care to fill in the rest of the proof?
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  3. #3
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    Hi,

    The proof is from Trench's Introduction to Real Analysis, which can be freely downloaded here

    William Trench - Trinity University Mathematics

    I was wary of posting the entire proof verbatim in case of any infringement (not sure what the laws are concerning mathematical proofs but I thought it better to be safe).

    The proof in question is on page 78 of the book.

    Thanks,

    Stonehambey

    EDIT: Actually I believe there may be a bracket missing from equation (10) in the proof, immediately after f'(g(x_0)). Can anyone confirm this?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Stonehambey View Post
    Hi,

    The proof is from Trench's Introduction to Real Analysis, which can be freely downloaded here

    William Trench - Trinity University Mathematics

    I was wary of posting the entire proof verbatim in case of any infringement (not sure what the laws are concerning mathematical proofs but I thought it better to be safe).

    The proof in question is on page 78 of the book.

    Thanks,

    Stonehambey

    EDIT: Actually I believe there may be a bracket missing from equation (10) in the proof, immediately after f'(g(x_0)). Can anyone confirm this?
    Ha! Luckily for you I own Trench, I don't think many people would download it . Anyways the part you mention reads


    \displaystyle \frac{h(x)-h(x_0)}{x-x_0}=\left[f'\left(g(x_0)+E(g(x))\right]\frac{g(x)-g(x_0)}{x-x_0}\right]



    when it should read


    \displaystyle \frac{h(x)-h(x_0)}{x-x_0}=\left[f'\overbrace{(g(x_0))}^{\text{here}}+E(g(x))\right]\frac{g(x)-g(x_0)}{x-x_0}\right]
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  5. #5
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    That makes a lot more sense now, thanks
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