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Math Help - Differentiable problem

  1. #1
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    Differentiable problem

    Suppose that f is continuous on (a,b), if f is also differentiable on (a,b), expect possibly at a single point x0 where lim(x--x0)(f'(x)) exists and is finite. prove that f is diff at x0 and that f' is cont at x0.

    can anyone solve this? i have no idea when i processed to lim(f'(x)).. thanks!
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  2. #2
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    Quote Originally Posted by shuxue View Post
    Suppose that f is continuous on (a,b), if f is also differentiable on (a,b), expect possibly at a single point x0 where lim(x--x0)(f'(x)) exists and is finite. prove that f is diff at x0 and that f' is cont at x0.

    can anyone solve this? i have no idea when i processed to lim(f'(x)).. thanks!

    f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} ; but since \lim_{x\to x_0}\left(f(x)-f(x_0)\right)'=\lim_{x\to x_0}f'(x)\,,\,\,\lim_{x\to x_0}(x-x_0)'=\lim_{x\to x_0}1 both exist and are finite, according to the given info, we can

    apply L'Hospital Rule and get \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x_0) , which exactly means that f'(x_0) exists and also that f'(x) is continuous at x_0 (these are definitions!)

    Tonio
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  3. #3
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    Some (including me) would prefer using directly the mean value theorem. It is really straightforward.

    Let us denote \ell=\lim_{x\to x_0}f'(x) (assumed to exist). Let \epsilon>0. Choose \delta>0 such that |x-x_0|<\delta and x\neq x_0 imply |f'(x)-\ell|<\epsilon. Then, for all such x, the mean value theorem provides c\in(x,x_0) such that f(x)-f(x_0)=f'(c)(x-x_0), hence \left|\frac{f(x)-f(x_0)}{x-x_0}-\ell\right|=|f'(c)-\ell|< \epsilon because |c-x_0|\leq |x-x_0|<\delta. This proves that f is differentiable at x_0 and f'(x_0)=\ell. The assumption f'(x)\to_{x\to x_0} \ell then reads " f' is continuous at x_0".
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