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This question already has an answer here:

Suppose we have a continuous function $f : \mathbb{R} \to \mathbb{R}$. Suppose also that for a certain point $c$, $\lim_{x \to c} f'(x)$ exists. Must $f'(c)$ exist as well, and be equal to this limit?

This isn't quite the same as asking if derivatives are always continuous. The well-known function $f(x) = x^2 \sin (1/x)$ is continuous and differentiable everywhere, but its derivative has no limit at $x = 0$. I'm wondering if the derivative of a continuous function can have a discontinuity where its limit does exist.

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marked as duplicate by Hans Lundmark, Jonas Dahlbæk, dantopa, Yujie Zha, Antonios-Alexandros Robotis Jun 29 '17 at 17:46

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For the limit to make sense, we have to assume that $f'$ exists on some interval around $c$.

If $\lim_{x\to c}f'(x)=L$, then $f'(c)$ exists and it is equal to $L$. Indeed, using the Mean Value Theorem we have $$ \frac{f(c+h)-f(c)}h=f'(\xi(h)) $$ for $\xi(h)$ between $c$ and $c+h$. As $h\to0$, $c+h\to c$ and so $\xi(h)\to c$. So $$ \lim_{h\to 0}\frac{f(c+h)-f(c)}h=\lim_{h\to0}f'(\xi(h))=L. $$

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  • $\begingroup$ I want to ask for the second equation, why the second equality holds. $\endgroup$ – Peter Liu Apr 28 '16 at 14:05
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For a quicker proof, use L'Hopital's rule

$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{f'(x)}{1}=\lim_{x\to a}f'(x)$$

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