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The mean value theorem says that for $f: [a,b], \rightarrow \mathbb{R}$ cont on [a,b] and diff on $(a,b)$, there exists a $\xi$ so $f'(\xi) = \frac{f(b) - f(a)}{ x - a}$.

Is it possible to replace $b$ with $x$ and let $x$ be a variable that approaches a, so that we instead get $$f'(\xi) = \frac{f(x) - f(a)}{ x - a}$$ where $\xi \rightarrow a$ as $x \rightarrow a$? (since $\xi \in (a,x)$)?

I want to prove differentiability in a end point, and so if we let $x \rightarrow 0$, we know that the RHS approaches (if it exists) the derivative $f'(a)$, so looking at the LHS, if we know the limit of $f'(\xi)$ as $\xi$ approaches $a$, then that means the limit is the derivative?

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  • $\begingroup$ Yes, your argument works. $f'(a)$ exists if and only if the limit $\lim_{x \to a} f'(x)$ exists, and they are the same. $\endgroup$ – Crostul Mar 1 '15 at 10:19
  • $\begingroup$ @Crostul I doubt if your assertion is correct? $\endgroup$ – user99914 Mar 1 '15 at 10:27
  • $\begingroup$ I guess $f(x) = x^2 \sin(1/x)$ when $x>0$ and $f(x) = 0$ when $x\le 0$ would be a counterexample. $\endgroup$ – user99914 Mar 1 '15 at 10:43
  • $\begingroup$ The real statement is if $\operatorname{lim}_{x \to a^{-}} f'(x) = l^-$ exists and $\operatorname{lim}_{x \to a^{+}} f'(x) = l^+$ exists, then $f$ differentiable (i.e. $f'(a)$ exists) $\Leftrightarrow l^+ = l^-$. The counterexample doesn't work in this case, as the limits do not exist; hence we need to go back to the difference quotient to prove the differentiability. $\endgroup$ – Jean-Luc Bouchot Mar 2 '15 at 20:16

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