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Let $f:[0,1) \rightarrow \mathbb{R}$ be continuous on [0,1) and differentiable on $(0,1)$. Suppose the limit $\lim_{x \to +0} f'(x)$ exists. Prove that $f$ is differentiable from the right at $0$.

My Attempt

Im not completely sure how to approach this; do I use MVT? i.e there exists $c \in (0,1)$ s.t $$f'(c)=\dfrac{f(1)-f(0)}{x-0}$$ and we know that $\lim_{x \to +0} f'(x)$ exists , so $$\lim_{x \to +0} \dfrac{f(1)-f(0)}{x-0}$$ which os equivalent to $f'(x)$ being differentiable from the right at 0.

Is this correct?

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    $\begingroup$ Your stetment of MVT is wrong. $\forall x\in [0,1)$ there exists a $c\in (0,1)$ such that $\frac{f(x)-f(0)}{x-0}=f'(c)$. Note that this point $c$ depends on $x$. $\endgroup$ – TZakrevskiy May 17 '14 at 8:40

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