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I'm trying to prove that there is no $f$ that is differentiable over $\mathbb{R}$ such as that $f'(x)=(1+|x|)^\frac{1}{x}$ for every $x\neq0$

Steps:

  1. Show that: $\lim_{x\to0^+}f'(x)=e$, $\lim_{x\to0^-}f'(x)=e^-1$

  2. Because the limits of both sides exist but are different, $x=0$ is a jump discontinuity

  3. According to Darboux's thoerem, $f'$ discontinuity points can only be of essential kind. Therefore, $f'$ that is described, can't be the derivative of any function over $\mathbb{R}$ when $x\neq0$

Is this proof correct?

EDIT: The limits calculation:

$\lim_{x\to0^+}f'(x)=\lim_{x\to0^+}(1+|x|)^\frac{1}{x}=\lim_{x\to0^+}(1+x)^\frac{1}{x} = e$

$\lim_{x\to0^-}f'(x)=\lim_{x\to0^-}(1+|x|)^\frac{1}{x}=\lim_{x\to0^-}(1-x)^\frac{1}{x} = \lim_{x\to0^-}e^{\ln(1-x)^\frac{1}{x}}=\lim_{x\to0^-}e^{\frac{1}{x}\cdot\ln(1-x)}$

Using $L'hopital$: $\lim_{x\to0^-}\frac{\ln(1-x)}{x} = \lim_{x\to0^-}\frac{-1}{1-x} = -1$. Since $e^x$ is continuous, $\lim_{x\to0^-}e^{\frac{1}{x}\cdot\ln(1-x)}$ is $e^-1$

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    how did you do step $1$? – 2017-01-20
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    @S.Peter it doesn't. I didn't pay attention that $f$ has to be differentiable at $0$. I think that your proof is correct. Though note that the second limit is $e^{-1}$. – 2017-01-20
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    My bad. I Added the way I calculated the limits – 2017-01-20

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