# Showing that if $\lim\limits_{x \to a} f'(x)=A$, then $f'(a)$ exists and equals $A$ [duplicate]

Let $f : [a; b] \to \mathbb{R}$ be continuous on $[a, b]$ and differentiable in $(a, b)$. Show that if $\lim\limits_{x \to a} f'(x)=A$, then $f'(a)$ exists and equals $A$.

I am completely stuck on it. Can somebody help me please? Thanks for your time.

## marked as duplicate by David Mitra, Lord_Farin, user1729, Amzoti, Noah SnyderApr 29 '13 at 12:29

• That's not true, unless you define $f'(a)$ as the right-sided derivative of $f$ at $a$. Otherwise, $f(x)=1$ for $x \geq 0$ and $f(x)=0$ for $x < 0$ is a counter-example. – fgp Apr 29 '13 at 11:53
• @fgp: No one is talking about right-sided limits here. – TMM Apr 29 '13 at 11:57
• @TMM Well, then $\lim_{x\to a} f'(x)$ is undefined, since $f'$ is undefined for $x < a$. – fgp Apr 29 '13 at 12:00
• Ah, for some reason I overlooked the $[a,b] \to \Bbb R$-part. But in any case, I don't see how your example is a counterexample. – TMM Apr 29 '13 at 12:02
• I was trying to make the point that one has to be a bit carefull about how you define $f'(a)$ if $f$ is only defined on $[a,b]$. I wanted to motivate the OP to think this through and state the required restrictions (like using the right-derivative) clearly in his question. So yeah, I was nitpicking, with a pedagogic intent ;-) – fgp Apr 29 '13 at 12:09

Let $\displaystyle{ \epsilon > 0}$. Since $\displaystyle{ \lim_{x \to a^+ } f'(x) = A }$ there exist $\displaystyle{ \delta >0 }$ such that for all $x$ with $a<x<a+ \delta$ is $\displaystyle{ |f'(x) -A| < \epsilon \quad (1)}$.

Let $x \in (a,a+ \delta)$ from Lagrange's Mean Value Theorem we get:

$$\frac{ f(x) - f(a) }{ x-a} = f'(c_x), \quad a < c_x < x < a+ \delta$$

Substitute in $(1)$ we get:

$$|f'(c_x) - A | < \epsilon$$

$\implies$

$$| \frac{f(x)-f(a)}{x-a} -A | < \epsilon$$

$\implies$

$$\lim_{ x \to a^+} \frac{f(x)-f(a)}{x-a}= A$$

$\implies$

$$f'_{+} (a) =A$$

Let $\epsilon>0$. We want to find a $\delta>0$ such that if $0\lt x-a\lt\delta$ then $\left|\dfrac{f(x)-f(a)}{x-a}-A\right|\lt\epsilon$. If $x\gt a$ then MVT tell us that $\dfrac{f(x)-f(a)}{x-a}=f'(c)$ for some $c\in[a,x]$.
Now use that $\displaystyle\lim_{c\to a^+}f'(c)=A$ to find that $\delta$.