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?