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Math Help - Differntiable problem

  1. #1
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    Differntiable problem

    Suppose that the function g: \mathbb {R} \rightarrow \mathbb {R} is differentiable at x_0=0. Also, suppose that for each natural number n, g( \frac {1}{n} ) = 0. Prove that g(0)=g'(0)=0

    I really don't know how to start this one, please give me some hints, thanks!
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  2. #2
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    It's been too long since I did proofs so I don't have the language, but the key point to me seems to be that for any tiny distance around x=0, there are an infinite number of points such that g(x)=0. That's because of the second assumption that g(\frac{1}{n})=0 for all those large natural numbers. Hope that helps.
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    Suppose that the function g: \mathbb {R} \rightarrow \mathbb {R} is differentiable at x_0=0. Also, suppose that for each natural number n, g( \frac {1}{n} ) = 0. Prove that g(0)=g'(0)=0

    I really don't know how to start this one, please give me some hints, thanks!
    What tbgh says shows that g(0)=0. Indeed, g is continuous at 0 and \frac{1}{n}\to_n 0, hence g(0)=\lim_n g\left(\frac{1}{n}\right)=0.

    Once we know g(0)=0, we have, by definition, g'(0)=\lim_{x\to 0, x\neq 0}\frac{g(x)}{x}, and this limit is assumed to exist because g is differentiable. Then it is the same as for the first step.
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