Show $f''(0)=0$ if $f(x_n)=0$ for $x_n \to 0$ Let $f$ be differentiable on an interval $A$ containing zero, and assume $(x_n)$ is a sequence in $A$ with $(x_n)\to 0$ and $x_n \neq 0$. Also, $f(x_n)=0$ for all $n$.
With this information I can show that $f(0)=0$ and $f'(0)=0$. (Namely, since $f$ is differentiable, it is continuous at $0$, so $f(0)=\lim (f(x_n))$. And $f'(0) =\lim_{x \to 0} f(x)/x = 0$.)
Now I'm supposed to show that, if $f$ is twice-differentiable at zero, then $f''(0) = 0$. So far I wrote out $f''(0) = \lim_{x \to 0} f'(x)/x$, and noticed that the fraction evaluates to $0/0$ at $0$. So in the spirit of the chapter, I thought I should use L'Hospital's Rules. However, evaluating 
$\lim_{x \to 0} f''(x)$ isn't much help here, because $f''(0)$ is exactly what I don't know...
Note: this is an exercise in Abbott in a chapter about the Mean Value Theorem and L'Hospital's Rules.
 A: Since $f(x_n)=f(0)=0$, Rolle theorem implies the existence of $y_n\in (0,x_n)$ such that $f^{'}(y_n)=f^{'}(0)=0$, Rolle again implies the existence of $z_n\in (0,y_n)$ such that $f^{''}(z_n)=0$. Since $\lim_nz_n=0, f^{''}(0)=\lim_nf^{''}(z_n)=0$. This proof is true if $f^{''}$ is continuous.
In the general situation, since $\lim_ny_n=0$ and $f^{'}(y_n)=0$, you have: $$\lim_{x\rightarrow 0}{{f'(x)-f'(0)}\over x}=\lim_n{{f'(y_n)-f'(0)}\over y_n}=0$$
A: Mean value Theorem implies the existence of $\{y_n\}$, such that
$$
0=0-0=f(x_n)-f(0)=(x_n-0)f'(y_n)
$$
and hence $f'(y_n)=0$ and $y_n\in(0,x_n)$, and thus $y_n\to 0$. Thus, assuming $f$ continuously differentiable at $0$,
$$
f'(0)=\lim_{n\to \infty}f'(y_n)=0.
$$
Similarly, MVT implies Mean value Theorem implies the existence of $\{z_n\}$, such that
$$
0=0-0=f(y_n)-f(0)=(y_n-0)f'(z_n)
$$
and hence $f'(z_n)=0$ and $z_n\in(0,y_n)$, and thus $z_n\to 0$. Thus, assuming $f$ twice continuously differentiable at $0$,
$$
f''(0)=\lim_{n\to \infty}f''(z_n)=0.
$$
