For a differentiable function, prove $f'(c) = \lim_{x \rightarrow c} f'(x)$

Question

Suppose $f: ℝ \rightarrow ℝ$ is differentiable. Show that if $c \in ℝ$ and $\lim_{x \rightarrow c} f'(x)$ exists, then $$f'(c) = \lim_{x \rightarrow c} f'(x)$$

The hint on this question is that derivatives have the Intermediate Value Property. Other than that, I really don't know where to start with this.

Once I get this started, I'm sure I'd be able to figure it out from there.

Answer 1

$\dfrac {f(x)-f(c)}{x-c}=f'(t_x)$, where

$t_x \in (\min (c,x), \max (c,x))$.

$\lim x \rightarrow c$ implies $\lim t_x \rightarrow c$.

$\lim_{x \rightarrow c} \dfrac{f(x)-f(c)}{x-c}=$

$\lim_{t_x \rightarrow c} f'(t_x) =L.$

$ \lim_{ x \rightarrow c} \dfrac{f(x)-f(c)}{x-c}$ exists and is equal to $L,$ $f$ is differentiable at $c.$

Answer 2

Assume the intermediate value property for derivatives. Let's pretend that your limit is $2$ and say for contradiction that $f'(c)=3$. (You can do the epsilon-delta version yourself).

Then for all $x$ in some interval $(c,c+\delta]$, the derivatives $f'(x)$ will be very close to $2$, so in particular all are less than 2.5. But this violates the intermediate value property applied to the intermediate value 2.7, say. That property says the derivative 2.7 should have been attained somewhere in $(c,c+\delta]$ since derivatives of 2 and 3 were attained around it.