Let $g(x)=e^{-1/x^2}$ for $x\not=0$ and $g(0)=0$. Show that $g^{(n)}(0)=0$ for all $n\in\Bbb N$. Let $g(x)=e^{-1/x^2}$ for $x\not=0$ and $g(0)=0$. Show that $g^{(n)}(0)=0$ for all $n\in\Bbb N$.
In the text it is already proven that for the function $f$ with $f(x)=e^{-\frac{1}{x}}$ for $x>0$ and $f(x)=0$ for $x\leq 0$, we have $f^{(n)}(0)=0$.
This is what we thought:
As $g(x)=f(x^2)$ for $x\in\Bbb R$, we may be able to use this. As $g^{(n)}$ becomes very hairy after taking a couple of derivatives. Any ideas ? 
 A: Hint: Prove recursively the stronger property that for every $n\geqslant0$ there exists a polynomial $P_n$ such that $g^{(n)}(x)=P_n(1/x)\,\mathrm e^{-1/x^2}$ for every $x\ne0$ and $g^{(n)}(0)=0$. 
The induction will use the fact (which you might want to prove) that, for every polynomial $Q$, $Q(1/x)\,\mathrm e^{-1/x^2}\to0$ when $x\to0$.
A: Using induction and L'Hôpital's rule it suffices to prove that for every $n\in\mathbb N$
$$
  \lim_{x\to 0}\frac{~~e^{-\frac{1}{x^2}}~~}{x^n}
~=~
  0
$$
We can assume that $n=2k$ is even, since if the assert holds for $n$ so it does for $n-1$. Therefore, setting $t=\frac{1}{x^2}$ we have to show that
$$
  \lim_{x\to +\infty}\frac{~~t^n~~}{e^t}
~=~
  0
$$
Applying $n$ times L'Hôpital's rule you have
$$
  \lim_{t\to +\infty}\frac{~~t^n~~}{e^t}
~=~
  \lim_{t\to +\infty}\frac{~~n!~~}{e^t}
~=~
  0
$$
EDIT
As commented by @Did the answer requires a bit more details; namely we are going to prove the following
If $h:\mathbb R\rightarrow\mathbb R$ is a function such that


*

*$h\in C^\infty(U_0\setminus\{0\})$, where $U_0$ is a neighborhood of $0$;

*$\displaystyle{\lim_{x\to 0}\frac{h(x)}{x^n}=0}$ $~\forall n\geq 0$,


then $h$ has derivatives of any order in $x=0$ and $h^{(n)}(0)=0$ $\forall n$.
Proof. Let us first check the assert for $n=1$:
$$
  h'(0)
~=~
  \lim_{x\to 0}
  {
    \frac{h(x)-h(0)}{x}
  }
~=~
  \lim_{x\to 0}
  {
    \frac{h(x)}{x}
  }
~=~
0
$$
and the last two equalities hold for the assumptions on $h$. Now, let us use induction. Suppose $n\geq 1$ and that the claim holds for every $k\leq n$, i.e. that $h$ in $0$ has derivatives of order up to $n$, and that $h^{(k)}(0)=0~$ $\forall k=0\ldots n$. Let us consider the following limit:
$$
  \lim_{x\to 0}
  {
    \frac{h(x)}{x^{n+1}}
  }
~=~
  0
$$
Since we know it exists and that $h$ has derivatives in $x=0$ up to the $n$-th, the first assumption on $h$ makes it possible to use L'Hôpital's rule $n$ times, reducing the previous equality to
$$
  \lim_{x\to 0}
  {
    \frac{h^{(n)}(x)}{(n+1)!x}
  }
~=~
  0
\quad\Rightarrow\quad
  \lim_{x\to 0}
  {
    \frac{h^{(n)}(x)}{x}
  }
~=~
  0
$$
Keeping in mind the second equality, we have that
$$
  h^{(n+1)}(0)
~=~
  \lim_{x\to 0}
  {
    \frac{h^{(n)}(x)-h^{(n)}(0)}{x}
  }
~=~
  \lim_{x\to 0}
  {
    \frac{h^{(n)}(x)}{x}
  }
~=~
0
$$
and this completes the proof.
