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Please prove: Let $f:(a,b) \to R$ be differentiable function, and let $c \in (a,b). $ If $\lim_{x \to c}f'(x)$ exists and is finite, then this limit must be $f'(c)$.

I tried doing it directly but it doesn't quite work out. I experimented on some common functions (like $\sin(x), x^2$) and it seems pretty obvious.

I'm thinking about using contradiction, although after writing down some definitions, I couldn't proceed. Or, probably if I could show that $f'(x)$ is continuous, then I win. Any idea? Thanks!

Alex
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3x89g2
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1 Answers1

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Hint: By the mean value theorem, there exists a $c_h \in (x, x + h)$ such that

$$\frac{f(x + h) - f(x)}{h} = f'(c_h)$$

  • Oh! Actually is there a way without using MVT? We haven't talked about it yet. – 3x89g2 May 28 '14 at 05:33
  • To be pedantic, you need to include $h$ negative, so you have to say $c_h\in(x-|h|,x+|h|)$. – Thomas Andrews May 28 '14 at 05:33
  • @ThomasAndrews Indeed that's true - I just wanted to give a brief hint answer that can easily be expanded to a full proof. –  May 28 '14 at 05:35
  • @Misakov I'd have to think about it for a little bit; the MVT approach is what first came to mind. What tools do you have available? –  May 28 '14 at 05:40
  • @user61527 This is my first undergraduate analysis course. We just finished differentiation (like definition and chain rule). Actually we're gonna talk about MVT next class, so probably I was just too hurried :p – 3x89g2 May 28 '14 at 06:18
  • That's interesting. I actually managed to prove it using intermediate value theorem for derivatives...mind explain how you would proceed using mean value theorem? – 3x89g2 May 29 '14 at 22:30
  • @Misakov Just take limits as $h \to 0$; by the squeeze theorem, $c_h \to c$. The left side is then $f'(c)$, while the right side is $\lim_{x \to c} f'(x)$. –  May 29 '14 at 22:49