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Math Help - Derivative exists

  1. #1
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    Derivative exists

    I have two functions. f: U->R is C^1 and f(0)=0. g: U->R defined with the piecewise function

    g(x)= {f(x)/x if x not equal to 0; and lim (h->0) of f(h)/h if x=0}

    I need to show that g(0) exists and that g is continuous (i.e. C^0).

    So I can see that g(0) implies that x=0 which means I need to show that the lim (h->0) of f(h)/h exists. I can see how f(h)/h came about, but i'm not sure how to find that it exists or that g is continuous. Thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by EricaMae View Post
    I have two functions. f: U->R is C^1 and f(0)=0. g: U->R defined with the piecewise function

    g(x)= {f(x)/x if x not equal to 0; and lim (h->0) of f(h)/h if x=0}

    I need to show that g(0) exists and that g is continuous (i.e. C^0).

    So I can see that g(0) implies that x=0 which means I need to show that the lim (h->0) of f(h)/h exists. I can see how f(h)/h came about, but i'm not sure how to find that it exists or that g is continuous. Thanks

    1. Let us first show that \lim_{\xi\to{0}}\frac{f(\xi)}{\xi} exists. We know that f(0)=0 which implies that

    \begin{aligned}\lim_{\xi\to{0}}\frac{f(\xi)}{\xi}&  =\lim_{\xi\to{0}}\frac{f(\xi)-0}{\xi-0}\\<br />
&=\lim_{\xi\to{0}}\frac{f(\xi)-f(0)}{x-0}\\<br />
&=f'(0)\end{aligned}

    Which exists since f\in\mathcal{C}^1

    2. Now since \lim_{\xi\to{0}}\frac{f(\xi)}{\xi} exists we can say that its left and right hand derivatives exist and equal f'(0). Or in other words \lim_{\xi\to{0}^-}\frac{f(\xi)}{\xi}=\lim_{x\to0^-}g(x)=f'(0)=\lim_{x\to{0}^+}\frac{f(\xi)}{\xi}=\li  m_{x\to0^+}g(\xi). Which implies that \lim_{\xi\to{0}}g(\xi) exists and equals f'(0).

    3. So \lim_{\xi\to{0}}\frac{f(\xi)}{\xi}=g(0)=f'(0)=\lim  _{\xi\to{0}}g(\xi) and we can finally conclude that g(0)=\lim_{\xi\to{0}}g(\xi) which is what was to be proven.


    Note: I assumed that 0 was a limit point of U but I think that was implied.
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