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Suppose $f(x)$ is continuous on $[a,b)$ and differentiable on $(a,b)$ and that $f '(x)$ tends to a finite limit $L$ as $x \to a^+$. Then $f(x)$ is right-differentiable at $x=a$ and $f '(a)=L$.

(epsilon-delta proof not needed).

This is a practice exam question.

I am having trouble translating this into a 'mathematical' statement. The MVT states that there exists $c$, $a\leq c\leq b$, such that:

$f'(c) = (f(b)-f(a))/(b-a)$

I suppose to prove that $f(x)$ is right differentiable at $x=a$, using the MVT, I need to somehow show that as $x \to a^+$, $f'(c)=f'(a)=L$ ??? Am I on the right track here? Can someone help me get started?

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    Apply MVT with $b=x$. As $c$ is sandwiched between $a$ and $x$, then as $x\to a+$ we also get that $c\to ?$.2012-06-02

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