I have problems with this exercise in the book Real Analysis of Miklós Laczkovich:

Let $f$ be continuous on $(a,b)$ and differentiable on $(a,b)\setminus \{c\}$, where $a<c<b$. Prove that if $\lim\limits_{x\to c}f'(x)=A$, where $A$ is finite. then $f$ is differentiable at $c$ and $f'(c)=A$

Since this problem is in the section of the mean value theorem, I have tried to attack it from that side, but I can not see what considerations to take to solve it, I would appreciate any help, thank you.

  • 1
    $\begingroup$ I cleaned up your limits a little bit to use the standard LaTeX structures for them; feel free to change back if you don't like them this way. $\endgroup$ Nov 26, 2020 at 6:46
  • $\begingroup$ perfect, thank you very much @Steven, I had no idea that they can be expressed that way $\endgroup$
    – Haus
    Nov 26, 2020 at 6:48
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    $\begingroup$ Essentially a duplicate of math.stackexchange.com/q/257907. $\endgroup$
    – Martin R
    Nov 26, 2020 at 8:23

1 Answer 1


First, let us look at the left-hand side. For all $x_1 \in (a,c)$, there exists $c_1 \in (x_1, c)$ such that $f(x_1) - f(c) = f'(c_1) (x_1 - c)$ by MVT. Note that this is so because $f$ is continuous on $[x_1, c]$. Therefore, $$ \lim_{x_1 \nearrow c} \dfrac{f(x_1) - f(c)}{x_1 - c} = \lim_{c_1 \nearrow c} f'(c_1) $$ since $x_1 \nearrow c$ implies $c_1 \nearrow c$ by the condition. Hence, we have $$ \lim_{x_1 \nearrow c} \dfrac{f(x_1) - f(c)}{x_1 - c} = A $$ Likewise, we can do a similar process for the case that, say, $x_2 \in (c,b)$.


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