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Math Help - Differentiability

  1. #1
    Senior Member I-Think's Avatar
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    Differentiability

    Let f(x) = cos(1/x) for x\neq0 and f(0) = 0. Is the function
    g(x) =\int_0^x f differentiable at 0?

    I say no
    My reasoning:

    \displaystyle\lim_{x\to{0}} \frac{g(x)-g(0)}{x-0}<br />

    \displaystyle\lim_{x\to{0}}\frac{\int_0^x cos(1/x)}{x}

    \int_0^x cos(1/x)=xcos(\frac{1}{x})-\int_0^x \frac{sin(\frac{1}{x})}{x}

    So

    \displaystyle\lim_{x\to{0}} \frac{\int_0^x fcos(1/x)}{x}

    =

    \displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}

    But cos(\frac{1}{x}) does not exist, therefore the limit
    \displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}

    does not exist, so g not differentiable at 0
    Correct?
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  2. #2
    Banned
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    Quote Originally Posted by I-Think View Post
    Let f(x) = cos(1/x) for x\neq0 and f(0) = 0. Is the function
    g(x) =\int_0^x f differentiable at 0?

    I say no
    My reasoning:

    \displaystyle\lim_{x\to{0}} \frac{g(x)-g(0)}{x-0}<br />

    \displaystyle\lim_{x\to{0}}\frac{\int_0^x cos(1/x)}{x}

    \int_0^x cos(1/x)=xcos(\frac{1}{x})-\int_0^x \frac{sin(\frac{1}{x})}{x}

    So

    \displaystyle\lim_{x\to{0}} \frac{\int_0^x fcos(1/x)}{x}

    =

    \displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}

    But cos(\frac{1}{x}) does not exist, therefore the limit
    \displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}

    does not exist, so g not differentiable at 0
    Correct?


    Looks fine and nice.

    Tonio
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