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Let $f:[a,b] \to \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Show that $\lim_{x \to a} f'(x) = A > \Rightarrow f'(a)$ exists and equals $A$

I am unable to think of any way to solve this problem

I tried using mean value theorem on $[a,x ]$ for $a < x < b$

$\exists \; c \in ]a,x[ \; \Rightarrow f'(c) = \frac{f(x)-f(a)}{x-a} \Rightarrow \lim_{x \to a} f'(c) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} $

Now the RHS represents $f'(a)$ but how does it equals $A$? because $f'(c) $ is a constant

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marked as duplicate by Hans Lundmark, Paramanand Singh, Arnaud D., Misha Lavrov, Claude Leibovici Dec 5 '17 at 5:47

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  • $\begingroup$ The LHS equals $A$ and therefore the RHS also equals $A$ and thus $f'(a) =A$. Note that $f'(c) $ is not a constant but by definition it depends on $x$. $\endgroup$ – Paramanand Singh Dec 2 '17 at 6:53
  • $\begingroup$ @ParamanandSingh How does LHS Equal A ? $\endgroup$ – So Lo Dec 2 '17 at 7:12
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    $\begingroup$ $c$ depends on the interval $[a, x]$. If $x$ changes the interval changes and thus $c$ also changes. To make thing concrete if $a=0$ then the value of $c$ for interval $[0,0.1]$ and for interval $[0,0.2]$ are not necessarily same. Don't you think so? And $c\to a$ as $x\to b$ and not $c\to b$ as $x\to b$. This happens because $a<c<x$ and when you take limit of this inequality as $x\to a$, $c$ must tend to $a$. $\endgroup$ – Paramanand Singh Dec 2 '17 at 11:59
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    $\begingroup$ If you want to deal with $b$ then the situation has to be different like $x<c<b$ and then let $x\to b$. In that case $c\to b$. $\endgroup$ – Paramanand Singh Dec 2 '17 at 12:02
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    $\begingroup$ There is a minor typo in my comment which I am fixing here. And $c\to a$ as $x\to\color{red} {a} $ and not $c\to b$ as $x\to b$. $\endgroup$ – Paramanand Singh Dec 2 '17 at 12:13
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You are very close. A slight rewrite will give you the result.

So take $x_n = a + \epsilon_n$, where $\epsilon_n > 0$ are so small that $a+\epsilon_n < b$ and $\epsilon_n \to 0$ as $n \to \infty$.

Apply the mean value theorem to $[a,x]$ to conclude that there is $c_n \in (a,x_n)$ such that $f'(c_n) = \frac{f(x)-f(a)}{x-a}$. Now, take limits on both sides : the right side is the derivative of $a$, and the right hand side is $\lim_{n \to \infty} f'(c_n)$, which exists and equals $A$. Hence, the result follows.

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