4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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 '17 at 13:22
2
$\begingroup$

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$$\lim_{n\in\mathbb N}f'(x_n)=\lim_{n\in\mathbb N}\frac{f\left(x_0+\frac1n\right)-f(x_0)}{\frac1n}=f'(x_0).$$

$\endgroup$
  • $\begingroup$ oh, thank you so much. it's so neat and clean and i wasted so much time going to the wrong directions. $\endgroup$ – BeginningMath Jul 1 '17 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.