# Analysis proof mean value theorem [duplicate]

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?

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

• 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??? – Danathon Apr 26 '13 at 4:28
• I guess you dropped some primes. – Sangchul Lee Apr 26 '13 at 4:36

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 $f'(c)=L$ .