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Math Help - Differentiability definition

  1. #1
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    Differentiability definition

    If  g:[-1,1] \rightarrow \mathbb{R} is differentiable with  g(0)=0 and  g(x) \neq 0 \forall x\neq 0 and f:\mathbb{R} \rightarrow \mathbb{R} is continuous with \frac{f(x)}{g(x)} \rightarrow 1 as x \rightarrow 0 then prove that f is differentiable at 0.

    So, from the definition we want to show that \lim_{x \to 0} \frac{f(x)-f(0)}{x} exists.

    From \lim_{x \to 0} \frac{f(x)}{g(x)} = 1 we have \lim_{x \to 0} f(x) = \lim_{x \to 0} g(x) = g(0) = 0 = f(0)

    Then,

    \lim_{x \to 0} \frac{f(x)-f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} -\lim_{x \to 0} \frac{f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{f(x)g(x)}{x(g(x))} = \lim_{x \to 0} \frac{f(x)}{g(x)} \cdot \lim_{x \to 0} \frac{g(x)}{x} = g'(0)
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  2. #2
    MHF Contributor

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    If you are asking if that is a valid proof, yes, it is. Very well done! The one comment I would make is that the middle term in
    \lim_{x\to 0}\frac{f(x)- f(0)}{x} = \lim_{x\to 0}\frac{f(x)}{x}- \frac{f(0)}{x} = \lim_{x\to 0} \frac{f(x)}{x}
    is not necessay. Just set f(0) in \frac{f(x)- f(0)}{x} equal to 0 to go directly:
    \lim_{x\to 0}\frac{f(x)- f(0)}{x} = \lim_{x\to 0} \frac{f(x)}{x}
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