Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f:\Bbb{R}\to\Bbb{R}$ such that

$$f(x) = \begin{cases} c_n & \text{if }x=\frac1n\text{ for some }n\in\mathbb N \\ 0 & \text{elsewhere} \\ \end{cases}$$
where $c_n$ is a given sequence. Find the condition on the sequence $c_n$ such that $f'(0)$ exists.

Here is what I got, I'm actually not sure only about the last part, when I choose the condition for my $N$.

So now, be definition of derivative, $f'(0)$ defined when $$\lim_{x\to0}{f(x)-f(0)\over x-0}=\lim_{x\to0}{f(x)\over x}$$ Now, if $x={1\over n}$ then ${f(x)\over x} = \begin{cases} nc_n & \text{for n =1,2,3...} \\ 0 & \text{elsewhere} \\ \end{cases} $

We have $$f'(0)=\lim_{x\to0}{f(x)\over x}=\lim_{n\to\infty}{f(1/n)\over 1/n}=lim_{n\to\infty}{c_n\over 1/n}=\lim_{n\to\infty}{nc_n}.$$

Now, $f'(0)$ exists only when $\lim_{n\to\infty}{nc_n}=0$. So this is our condition. Let $\epsilon>0$, need to find an $N>0$ such that $$|nc_n-0|<\epsilon \text{ for }n>N$$ $\rightarrow |c_n|<{\epsilon\over |n|}$

Let $N={\epsilon\over |n|}$ so that we have $$|nc_n|<|n|{\epsilon\over |n|}=\epsilon$$

So $\lim_{n\to\infty}nc_n=0$ hence, $f$ is differentiable at $0$. And we have it (?)

What do you think?

share|improve this question
I think your accept rate is painfully low. I'd enhance it at once. –  DonAntonio Nov 16 '12 at 19:58
what do you mean by accept rate? –  Akaichan Nov 16 '12 at 20:00
Read it under your nick,@user45593. You have accpet rate of 25%, which means you've only accepted as "the best answer" about one quarter of the questions you've asked, and this could mean you don't really like the answers you get here. –  DonAntonio Nov 16 '12 at 20:02
Oh, ok, thanks. And I do like the answer I get here. And sometimes, I answered question myself also, just like this one. –  Akaichan Nov 16 '12 at 20:04
You "wrote something wrong" at some point : when you write : when $x = 1/n$, $f(x) = \{ nc_n,$ if $x = 1/n$ for $n = 1,2, \dots$. It should be written $c_n$ instead of $nc_n$. And you could've stopped at "So this is our condition." The rest is irrelevant. But the argument is fine –  Patrick Da Silva Nov 16 '12 at 20:07

1 Answer 1

Most books define differentiable functions on an interval $ (a, b) $. But we can define the derivative of a function $ f: A \to \mathbb {R} $ at a point $ c \in A $ if is $ c $ is an accumulation point ( or cluster point ) of $ A $.

Then $f^\prime(c)$ is unic number that have the propert:

$$ \forall \epsilon >0, \exists \delta=\delta(c,\epsilon)>0\quad \mbox{ such that } x\in A \mbox{ e }0<|c-x|<\delta \implies \left|\frac{f(x)-f(c)}{x-c}- f^\prime(c) \right|<\epsilon. $$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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