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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.

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2 Answers 2

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

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  • $\begingroup$ Might be better to write $t_x$ in place of $t.$ $\endgroup$
    – zhw.
    Nov 1, 2018 at 16:10
  • $\begingroup$ zhw.Thanks for your suggestion . $\endgroup$ Nov 1, 2018 at 16:36
  • $\begingroup$ Thank you so much! $\endgroup$
    – user610107
    Nov 1, 2018 at 17:18
  • $\begingroup$ user610107.A pleasure.This question has been asked before on this site.You may find other answers dealing with this problem. $\endgroup$ Nov 1, 2018 at 18:01
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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.

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  • $\begingroup$ But this doesn't show $f'(c)$ exists. $\endgroup$
    – zhw.
    Nov 1, 2018 at 16:09
  • $\begingroup$ $f$ is differentiable (see problem statement) $\endgroup$
    – Matthew C
    Nov 1, 2018 at 17:23
  • $\begingroup$ But I agree that the assumption is unnecessary as the other proof shows. $\endgroup$
    – Matthew C
    Nov 1, 2018 at 17:26
  • $\begingroup$ Yes, you're right. I had assumed it was the more general result. $\endgroup$
    – zhw.
    Nov 1, 2018 at 17:28

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