$f(\frac{1}{n})=1$ for all positive integers $n$. Prove that $f'(0)=0$ and hence prove that $f''(0)=0$ $f:(-1,1) \rightarrow \mathbb{R}$ be a function double differentiable throughout.$f(\frac{1}{n})=1$ for all positive integers $n$.
Prove that $f'(0)=0$ and hence prove that $f''(0)=0$
My take:
We know $f'(0)= \lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h} =\lim_{n \rightarrow \infty} \frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}}=\lim_{n \rightarrow \infty } n(1-f(0))$
From here I don't know how to proceed from here.
Please help
 A: Since $f$ is twice differentiable then $f$ is continuous.  We have immediately $f(0)=1$ since $\lim_{n\to \infty}f(1/n)=f(0)=1$.

From the mean value theorem, there exists $\xi_n\in (0,1/n)$ such that
$$f'(\xi_n)=\frac{f(1/n)-f(0)}{ \frac1n} =0$$
Since $f$ is twice differentiable, the $f'$ is continuous and we have
$$0=\lim_{n\to \infty}f'(\xi_n)=f'(0)$$
as was to be shown.

From the mean value theorem, there exists $\eta_n\in (-1/n,0)$ such that
$$f(-1/n)=1-\frac1n f'(\eta_n)$$
Finally, we see that
$$\begin{align}
\lim_{n\to \infty}\frac{f(1/n)-2f(0)+f(-1/n)}{1/n^2}&=\lim_{n\to \infty}\frac{(1)-2(1)+\left(1-\frac1n f'(\eta_n)\right)}{1/n^2}\\\\
&=- \lim_{n\to \infty}nf'(\eta_n)\\\\
\end{align}$$
But using Taylor's Theorem with the Peano form of the remainder, $f'(\eta_n)=o(\eta_n)$, which implies that $nf'(\eta_n)\to 0$ as $n\to\infty$.
Hence, we find $f''(0)=0$ as was to be shown!.
A: How about this solution?
Since,$f$ is differentiable at $x=0$,
$$f'(0)=\lim_{n\to \infty} \frac{ f(\frac 1n)-f(0) }{\frac 1n} =0$$.
Now by Taylor's theorem we have, $f(\frac 1n)=f(0)+f'(0)\cdot \frac 1n +f''(c_n)\cdot \frac{1}{2n^2}$
$\implies f''(c_n)=0$ where $c_n\in (0,\frac 1n)$
As $n\to \infty$,$c_n \to 0$.
So, $f''(0)=0$.(I think here we needed an extra assumption that f'' is continuous). Do you have any way to overcome this problem?
