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. :)

  • $\begingroup$ $n(f(x_0+1/n)-f(x_0))=\frac{f(x_0+1/n)-f(x_0)}{1/n}$. Use the definition of $f'(x_0)$. $\endgroup$ – Reveillark Mar 15 at 15:43
  • $\begingroup$ Hello Janka, please use mathjax for formulas. Hint: look up the definition of the derivative. $\endgroup$ – maxmilgram Mar 15 at 15:44
  • $\begingroup$ 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. $\endgroup$ – Janka Mar 15 at 16:19
  • $\begingroup$ And how can I find out, if the sequence is convergent ? $\endgroup$ – 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?

  • $\begingroup$ I still didn't get it, that "intersteps" $\endgroup$ – Janka Mar 15 at 17:16

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