function with zero derivatives at certain points Let $f:\mathbb{R}_{\ge0}\rightarrow \mathbb{R}$ be a $\mathcal{C}^{\infty}$ function such that $\lim_{x\rightarrow \infty}f(x) = 0$. If $f(0)=0$, I know how to show that $f'(c)=0$ for some $c$, and this is intuitively clear. My question is :
Is it possible to find a sequence $x_n$ such that $f^{(n)}(x_n)=0$ for every $n$ ?
It looks like a very strong statement, but I feel like once it is true for $f'$ it is also true for all the other derivatives of $f$ at some point since we can just "repeat" by induction. I am not able to construct a proof though...
 A: This is indeed true. The existence of $x_1$ such that $f'(x_1)=0$ is a direct consequence of the fact that since $f(0)=0$ and $f(x)\overset{x\rightarrow +\infty}\longrightarrow 0$, $f$ cannot be monotone which implies that $f'$ vanishes at some point.
I am going to construct $(x_n)$ by induction as suggested, and the statement I am going to prove is a bit stronger : I claim that a can construct a sequence $(x_n)$ that is strictly increasing and that verifies your condition.
Suppose we have constructed the sequence up to $n$, and suppose that $f^{(n+1)}(x)\neq 0$ for any $x>x_n$. The continuity of $f^{(n+1)}$ implies that the sign of $f^{(n+1)}$ is constant on $(x_n,+\infty)$, and assume that it is positive. Therefore, $f^{(n)}$ is strictly increasing on $[x_n,+\infty]$. Let $a>x_n$.
Using Taylor expansion, for any $x\geq a$, there exists $y\in[a,x] such that
$$f(x)=f(a)+(x-a)f'(a)+...+\frac{(x-a)^{n-1}}{(n-1)!}f^{(n-1)}(a)+ \frac{(x-a)^{n}}{n!}f^{(n)}(y),$$ which in turn implies that
$$f(x)\geq \underbrace{f(a)+(x-a)f'(a)+...+\frac{(x-a)^{n-1}}{(n-1)!}f^{(n-1)}(a)+ \frac{(x-a)^{n}}{n!}f^{(n)}(a)}_{:=g(x)},$$
since $f^{(n)}(y)\geq f^{(n)}(a)$ because $f^{(n)}$ is increasing.
By the induction hypothesis, $f^{(n)}(x_n)=0$, hence $$f^{(n)}(a)>f^{(n)}(x_n)=0.$$
This implies that $g(x)\overset{x\rightarrow +\infty}{\longrightarrow} +\infty$. Therefore, it is also the case for $f(x)$ since $f(x)>g(x)$ on $[x_n,+\infty]$. This contradicts the hypothesis on $f$. We conclude that there exists $x_{n+1}>x_n$ such that $f^{(n+1)}(x_{n+1}) = 0$, which is what we wanted to prove.
EDIT : Someone suggested that by Baire Category Theorem, having derivatives that vanishes at some points implies that the function is a polynomial. This interpretation of the theorem is wrong, and I want to clarify it for the OP of the question. The theorem says that if there exists a rank $N$ such that the $n$-th derivatives are IDENTICALLY ZERO for any $n\geq N$, then the function is a polynomial. This is not the same as saying that all the derivatives vanish at some point. It has also been suggested that the function $xe^{-x}$ was a counterexample, which is not true. Indeed, a quick computation show that
$$\frac{d^n}{dx^n}xe^{-x} = (-1)^n e^{-x}(x-n),$$ which vanishes at $n$.
