# Is that statement true or false?

Are following statements true or false ?

If function $$f$$ is differentiable at $$x_0$$, then the sequence n.$$( f(x_0+(1/n)) - f(x_0))_{n\in\mathbb{N}}$$ is convergent.

I am really not sure, but I think it is true, but just because it seems to me so and I can't imagine any proof of it, would you help me ? Maybe it has something with Heine's sentence (a transition from function to sequence) but I see nothing in it. Thanks for your help. :)

• $n(f(x_0+1/n)-f(x_0))=\frac{f(x_0+1/n)-f(x_0)}{1/n}$. Use the definition of $f'(x_0)$. – Reveillark Mar 15 at 15:43
• Hello Janka, please use mathjax for formulas. Hint: look up the definition of the derivative. – maxmilgram Mar 15 at 15:44
• And what should I do with that definition ? I am not very smart in derivations because we started with them only few days ago so I don't know what to do. – Janka Mar 15 at 16:19
• And how can I find out, if the sequence is convergent ? – Janka Mar 15 at 17:47

Dfferentiability at $$x_0$$ implies continuity at $$x_0$$. In particular, $$\lim_n f\left(x_0+\frac{1}{n}\right) - f(x_0) = \lim_n \frac{f\left(x_0+\frac{1}{n}\right) - f(x_0)}{\frac{1}{n}} \cdot \frac{1}{n}.$$ You can evaluate each limit separately (limit of product is product of limits) to get $$\lim_n \cdot = f^\prime(x_0) \cdot 0 =0$$. So, yes the sequence converges (to zero).
Edit: It seems that I missed the $$n$$ in front. In that case, $$\lim_n n\cdot \left(f\left(x_0+\frac{1}{n}\right) - f(x_0)\right) = \lim_n \frac{f\left(x_0+\frac{1}{n}\right) - f(x_0)}{\frac{1}{n}} = f^\prime(x_0).$$ Now that I've addressed the right question, does his make sense?