# If $\lim_{x\to c}f'(x)=A$ then $f'(c)=A$ [duplicate]

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 $$a. Prove that if $$\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.

• 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. Nov 26, 2020 at 6:46
• perfect, thank you very much @Steven, I had no idea that they can be expressed that way
– Haus
Nov 26, 2020 at 6:48
• Essentially a duplicate of math.stackexchange.com/q/257907. Nov 26, 2020 at 8:23

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, $$x_2 \in (c,b)$$.