Let $f: [a,b] \to \mathbb{R}$ be such that $f', f''$ exist for every $x \in [a,b]$. Let $(x_n)$ be a sequence of distinct points in $[a,b]$ such that $$\lim_{n \to \infty} x_n=y$$ and
$$f(x_n)=0$$ for every $n \in \mathbb{N}$. Prove that $f(y)=f'(y)=f''(y)=0$.
I don't know why they stipulate $f$ is twice-differentiable for every $x \in [a,b]$. Isn't it meaningless to define differentiability at a boundary?
Anyway. My reasoning was the following.
(1) We know $y \in [a,b]$ because closed sets contain their boundary points. Hence, $f$ is differentiable at $y$ and therefore also continuous at $y$, which means
$$\lim_{x \to y}f(x) = f(y) \tag{i}$$
But if this limit exists, we know that it's preserved under composition with any sequence such that $s_n \to y$; in particular,
$$\lim_{n \to \infty} f(x_n)= \lim_{x \to y}f(x) \tag{ii}$$
But $f(x_n)\equiv 0$, so (i) and (ii) jointly imply $f(y)=0$.
(2) Since $f'(y)$ exists, by definition the limit
$$\lim_{x \to y} \frac{f(x) - f(y)}{x - y} = \lim_{x \to y} \frac{f(x) - 0}{x - y} = f'(y)$$
must exist and be preserved under composition... etc. Since $f(x_n)\equiv 0$ (and the denominator doesn't vanish because we can take infinite $x_n$ such that $x_n \neq y$), the limit again must be zero. Hence $f'(y)=0$.
(3) By definition we have
$$f''(y) := \lim_{x \to y} \frac{f'(x) - f'(y)}{x - y} \stackrel{\text{L'Hôpital}}{=} \lim_{x \to y} f''(x)$$
so $f''$ must be continuous at $y$. If $f''(y)$ didn't vanish, its continuity would imply that there's an interval $I^* = (y - \epsilon, y + \epsilon)$ in which it preserves its sign. Notice that then $f'$ would be strictly monotonous in $I^*$. Since $f(y)=f'(y)=0$, $f$ would be strictly monotonous in $(y - \epsilon, y)$ and in $(y, y+\epsilon)$.
But this is absurd because there are infinite points $x_n$ in $I^* - \{ y \}$ such that $f(x_n)=0$. Hence, $f''(y)$ must vanish.
Is my proof correct?
