If $f$ is twice differentiable such that $f\left(\frac{1}{n}\right)=0$, then what can we say about $f''(0)$? The continuity of $f, f'$ and Rolles theorem implies that $f(0)=0$ and $f'(0)=0$, but is it true that $f''(0)$. We have a sequence $(x_n)$ such that $f''(x_n)=0$ and $x_n \to 0$. But continuity of $f''$ is not guaranteed. Is it true that $f''(0)=0$? Or can anyone give me a counterexample?
 A: The canonical example of a function with discontinuous derivative is $g(x) = x^2 \sin(1/x)$ for $x\neq 0$ and $g(0) = 0.$ One can check that for $x \neq 0$ we have $g'(x) = 2x \sin(1/x) - \cos(1/x)$ and $g'(0) = 0.$ It's easy to check that $g'$ oscillates around the origin and takes the value $-1/2$ infinitely often. Picking $f(x) = \int_0^x g(t) \, dt + x^2/4,$ we see that $f''(x) = g'(x) + 1/2,$ so since $g'(x) = -1/2$ happens infinitely often near the origin, we find $x_n\rightarrow 0$ with $g'(x_n) = -1/2$ and so $f''(x_n) = 0,$ despite the fact that $f''(0) = g'(0) + 1/2 = 1/2 \neq 0.$
A: $$ f\left(\frac{1}{n}\right)=f(0)+\frac{f'(0)}{n}+\frac{f''(0)}{2n^2}+o\left(\frac{1}{n^2}\right) $$
Therefore, replacing what needs to be replaced, we have
$$ \frac{f''(0)}{2n^2}=o\left(\frac{1}{n^2}\right) $$
Multiplying by $n^2$ on both sides gives that $f''(0)=o(1)$, by letting $n\rightarrow +\infty$ we have $f''(0)=0$. If you suppose $f$ to be $\mathcal{C}^{\infty}$, you can samely prove that $f^{(p)}(0)=0$ for all $p$ by induction.
