# Show that $\lim_{x \to a} f'(x) = A \Rightarrow f'(a)$ exists and equals $A$ [duplicate]

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

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

• 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 $x$. – Paramanand Singh Dec 2 '17 at 6:53
• @ParamanandSingh How does LHS Equal A ? – So Lo Dec 2 '17 at 7:12
• $c$ 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 $a$. – Paramanand Singh Dec 2 '17 at 11:59
• 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 $c\to b$. – Paramanand Singh Dec 2 '17 at 12:02
• 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 $x\to b$. – Paramanand Singh Dec 2 '17 at 12:13

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.