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Let $f$ be a function which is continuous on an open interval $I$. Let $c$ be a point of $I$. Suppose that $f$ is differentiable at every point of $I$ other than $c$, and that $\displaystyle\lim_{x\to c} f'(x)$ exists. Let $L$ denote this limit. Prove that $f$ is differentiable at $c$, and that $f'(c) = L$.

If $e > 0$ is given, choose a $\delta > 0$ such that $|f'(x)-L|<e$ whenever $e \in I$ and $0<|x-c|<\delta$. Now I need to apply the mean value theorem. Can someone help?

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marked as duplicate by Hans Lundmark, Jonas Dahlbæk, Frits Veerman, Trevor Gunn, Frpzzd Jun 29 '17 at 15:01

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  • $\begingroup$ If e > 0 is given, choose a ∂> 0 such that |ƒ’(x)-L|<e whenever e ϵ I and 0<|x-c|<∂. Now I need to apply the mean value theorem…can someone help??? $\endgroup$ – Danathon Apr 26 '13 at 4:28
  • $\begingroup$ I guess you dropped some primes. $\endgroup$ – Sangchul Lee Apr 26 '13 at 4:36
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We have $\displaystyle\lim_{x\to c} f'(x)=L$ , so there exists an interval $c\in (a,b)$ such

that if $x\in J=(a,c)\cup(c,b)$ then $L-\varepsilon <f'(x)<L+\varepsilon$ .

Now , if we take any $x_0 \in J$ , then by MVT , $L-\varepsilon <\frac{f(c)-f(x_0)}{c-x_0}=f'(x_1)<L+\varepsilon$

where $x_1\in J$ , too.

Since we can do this for every $\varepsilon >0$ , we conclude that $f'(c)=L$ .

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  • $\begingroup$ Thanks, that helped a lot!!! $\endgroup$ – Danathon Apr 26 '13 at 13:07

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