I'm having trouble with this one, I don't know what is the right path to solve it.

$f$ is a differentiable function in $(a,b)$, and let $x_0 \in (a,b)$. prove that there exists a sequence $(x_n)$ so that: $x_n \neq x_0$ for every $n$, $\lim_{n->\infty} x_n=x_0$ and $\lim_{n->\infty} f'(x_n) = f'(x_0)$

A hint I have: for every natural $n$, regard $[x_0,x_0+1/n]$ and use Lagrange's mean value theorem.

I tried to use the property and definition of a derivative to prove that $\lim_{n->\infty} f'(x_n) = f'(x_0)$, but couldn't find a way to conclude that $\lim_{n->\infty} x_n=x_0$. I could assume that, but I don't know how to conclude that or how to prove it.

Hoping you could help me. Thank you very much!

  • $\begingroup$ What does the MVT say about $(f(x_0+1/n)-f(x_0))(1/n)$ and what does this go to as $n\to \infty?$ (Also, you want to say it's a sequence, not a series) $\endgroup$
    – zhw.
    Jul 1, 2017 at 13:22

1 Answer 1


For each natural $n$, choose $x_n\in\left(x_0,x_0+\frac1n\right)$ such that $$f'(x_n)=\frac{f\left(x_0+\frac1n\right)-f(x_0)}{\frac1n}.$$Then $\lim_{n\in\mathbb N}x_n=x_0$. Besides$$\lim_{n\in\mathbb N}f'(x_n)=\lim_{n\in\mathbb N}\frac{f\left(x_0+\frac1n\right)-f(x_0)}{\frac1n}=f'(x_0).$$

  • $\begingroup$ oh, thank you so much. it's so neat and clean and i wasted so much time going to the wrong directions. $\endgroup$ Jul 1, 2017 at 13:32

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