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I got a problem with this theorem:

Theorem: Let $f:\;(a,b)\rightarrow\mathbb{R}$ be a continuous function. Suppose $c\in(a,b)$ is such that f is differentiable in $(a,c)$and in $(c,b)$ and $\lim_{x\rightarrow c}f'(x)=L$. Prove $f$ is differentiable in $c$ and $f'(c)=L$

Proof:

$$\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=\lim_{x\rightarrow c}\frac{f(c)-f(x)}{c-x}=\lim_{x\rightarrow c}f'(x)=L$$ Then $f'(c)$ exists and $f'(c)=L$

Is my proof fine?

String
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rcoder
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2 Answers2

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I will prove $\lim\limits_{t \to c^+} \frac{f(t)-f(c)}{t-c}=L$ and leave the left-handed limit case for you.

Take $\epsilon>0$. Now, consider $\delta>0$ such that $x-c<\delta \implies |f'(x)-L| < \epsilon.$

Now, if $x-c<\delta$, we have that $|\frac{f(x)-f(c)}{x-c}-L|=|f'(\xi)-L|<\epsilon,$ where the equality is given for some $\xi$ between $c$ and $x$ by the mean value theorem, and the inequality by the way we defined $\delta$. It follows that $\lim\limits_{t \to c^+} \frac{f(t)-f(c)}{t-c}=L$.

Aloizio Macedo
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Hints:

  • For $x$ sufficiently close to $c$ we have $f'(x)$ bounded in an interval based on $L\pm\varepsilon$.
  • This implies that $f(c)$ must be found between the lines with slopes $L-\varepsilon$ and $L+\varepsilon$ through $(x,f(x))$ for $x$ sufficiently close to $c$. Here the continuity of $f$ should play a role.
  • Since this can be achieved for all $\varepsilon>0$ the conclusion should follow.

Perhaps someone else will provide a completely different approach, but this was my intuition regarding this problem.

String
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  • @Bvss12: No problem, but I think the way Alozio Macedo has answered it is more of a standard approach, since some way of applying the mean value theorem will be the way to proceed. – String Oct 18 '16 at 09:20