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Math Help - g'(0) exists s.t. lg(x)l <or= (x^2) proof

  1. #1
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    g'(0) exists s.t. lg(x)l <or= (x^2) proof

    this question has two parts. i have tried to answer part a) but I don't think my proofs are very good. I'm practicing for an exam so if someone could explain the concepts I am supposed to have gotten from the question, I would very much appreciate it.

    note: a hint to part a) is that the previous question (which i solved) was to prove f is differentiable at 0 if f(x)=(x^2) for rational x and f(x)=0 for irrational x. I don't think I fully understand the connection to this problem--again, any conceptual help would be much appreciated.

    a) let g be a function such that lg(x)l <or= (x^2) for all x. Prove that g is differentiable at 0.

    b) this result can be generalized if (x^2) is replaced by lf(x)l where f has what property?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by 234578 View Post
    this question has two parts. i have tried to answer part a) but I don't think my proofs are very good. I'm practicing for an exam so if someone could explain the concepts I am supposed to have gotten from the question, I would very much appreciate it.

    note: a hint to part a) is that the previous question (which i solved) was to prove f is differentiable at 0 if f(x)=(x^2) for rational x and f(x)=0 for irrational x. I don't think I fully understand the connection to this problem--again, any conceptual help would be much appreciated.

    a) let g be a function such that lg(x)l <or= (x^2) for all x. Prove that g is differentiable at 0.

    b) this result can be generalized if (x^2) is replaced by lf(x)l where f has what property?
    Hint:
    Spoiler:


    a) |g(x)|\leqslant x^2\implies -x^2\leqslant g(x)\leqslant x^2\implies -x\leqslant\frac{g(x)}{x}\leqslant x

    b) \frac{f(x)}{x}\underset{x\to0}{\sim}0
    [/quote]
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