1. ## derivatives

If $\{x_n\}$and $\{y_n\}$ are sequences converging to the same point $x_0$ such that $x_n$ $<$ $x_0$ $<$ $y_n$ $\vee$ $n$, and if $f$is differentiable funtion whose derivative is continuous at $x_0$ and which satisfies the condition that $f(x_n)=f(y_n)$ $\vee$ $n$, show that $f'(x_0)=0$.

2. Originally Posted by Kat-M
If $\{x_n\}$and $\{y_n\}$ are sequences converging to the same point $x_0$ such that $x_n$ $<$ $x_0$ $<$ $y_n$ $\vee$ $n$, and if $f$is differentiable funtion whose derivative is continuous at $x_0$ and which satisfies the condition that $f(x_n)=f(y_n)$ $\vee$ $n$, show that $f'(x_0)=0$.
Does $x_n actually mean $x_n? If so, I would advise you to use the mean value theorem that guarantees $\forall n$ the existence of a $\xi_n$ such that $x_n <\xi_n and

$f'(\xi_n)\cdot(y_n-x_n)=f(y_n)-f(x_n)=0$

Hence $f'(\xi_n)=0, \forall n$ and by continuity of $f'$ at $x_0$ you conclude...

3. yes i meant $\forall n$ but how can you say $\xi_n$= $x_0$?

4. Originally Posted by Kat-M
yes i meant $\forall n$ but how can you say $\xi_n$= $x_0$?
Aw, I didn't say that. But what I can say is that $\lim_{n\rightarrow\infty}\xi_n=x_0$ (why? - because the sequence $\xi_n$ is sandwiched between two sequences that both converge to $x_0$). And thus, we have, by continuity of $f'$ at $x_0$:

$f'(x_0)=f'(\lim_{n\rightarrow \infty}\xi_n)=\lim_{n\rightarrow\infty}f'(\xi_n)=\ lim_{n\rightarrow\infty}0=0$