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Assume that $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0,$ $f'(x)$ exists. If $\lim_{x\rightarrow 0} f'(x)$ exists, does it follow that $f'(0)$ exists? Prove or disprove.

Intuitively, it seems to me that it does does. That $f'(x)$ would not exist implies a point discontinuity, since $\lim_{x\rightarrow 0} f'(x)$ exists. Given that $f$ is continuous, $f$ is defined at $x.$ However, I'm not sure how to go about proving this (if I'm correct). Any hints? Thanks :)

Rafael Vergnaud
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1 Answers1

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Yes, you are right! Just use Lagrange Mean Value theorem. $$f'(0)=\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\dfrac{f'(\xi)(x-0)}{x-0} =\lim_{x\to 0}f'(\xi)=\lim_{\xi\to 0}f'(\xi)=\lim_{x\to 0}f'(x),$$ here $\xi$ is between $0$ and $x$, and $x\to 0$ implies $\xi\to 0.$

Riemann
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  • Hey, Riemann. Is there any other way to show this? My book has not mentioned the Lagrange Mean Value Theorem. It has mention the mean value theorem and the ratio mean value theorem. – Rafael Vergnaud Nov 08 '18 at 03:13
  • what is the mean value theorem and the ratio mean value theorem in your book? – Riemann Nov 08 '18 at 03:15
  • Roll mean value theorem ,Lagrange mean value theorem and Cauchy mean value theorem are all called mean value theorem. – Riemann Nov 08 '18 at 03:17
  • see https://en.wikipedia.org/wiki/Mean_value_theorem – Riemann Nov 08 '18 at 03:18
  • MVT: A continuous function $f:[a,b] \rightarrow \mathbb{R}$ that is differentiable on the interval (a, b) has the following property: $f(b) - f(a) = f'(c)(b-a)$ for some $c \in (a, b)$ – Rafael Vergnaud Nov 08 '18 at 03:19
  • Yes， this is the Lagrange mean value theorem! $f(x)-f(0)=f'(\xi)x.$ – Riemann Nov 08 '18 at 03:20