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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 $f'(c)=L$ .