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

Solution 1:

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...

Can you complete the proof now?

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?