# $\lim_{x\to c}f'(x)=L$ implies $f'(c)=L$ [duplicate]

Let $f:[a,b]\to\mathbb{R}$ be differentiable on $[a,b]$ and let $c \in(a,b)$. Suppose that $\lim_{x\to c}f'(x)=L$ some $L \in\mathbb{R}$. Without using L'Hospital's Rule, prove that $f'(c)=L$.
Hint: Use the Mean Value Theorem and the e-d definition of f'(c).

Any leads would be much appreaciated. Thanks.

• What is wrong with the hint you already have? Did you try to use it? Where did you get stuck? – Dennis Gulko Feb 6 '13 at 14:02
• See this post for ideas. – David Mitra Feb 6 '13 at 14:03
• @DavidMitra i'm not sure where i have to use the epsilon-delta definiton of f'(c) here. – tal Feb 6 '13 at 15:32

Let $x>c$. By the Mean Value Theorem in $[c,x]$, $$\exists \xi_x\in [c,x]\text{ so that }f'(\xi_x)=\frac{f(x)-f(c)}{x-c}$$ Now as $x\to c^+$, $\xi_x\to c^+$ and so...
EDIT: With $\epsilon-\delta$: Let $\epsilon>0$. Then $$\exists \delta>0\text{ so that }c<y<c+\delta\implies \left|f'(y)-L\right|<\epsilon$$ As we saw above, for $x\in (c,c+\delta)$, $\exists \ y\in (c,x)\subseteq (c,c+\delta)$ so that $$f'(y)=\frac{f(x)-f(c)}{x-c}$$ But then, $$\left|\frac{f(x)-f(c)}{x-c}-L\right|<\epsilon$$ whenever $c<x<c+\delta$. Is it clear now?