Quotient of two smooth functions is smooth 
Let $f:\mathbb R\to \mathbb R$ be a $C^\infty$-smooth function. Suppose that   $f^{(k)}(0)=0$ for $k=0,\dots,n-1$. Prove that the function $g(x)=f(x)/x^n$ extends to a $C^\infty$-smooth function on $\mathbb R$. 

Comment: by l'Hôpital's rule, $g$ has a finite limit at $0$, namely $f^{(n)}(0)/n!$. So, it extends to a continuous function on $\mathbb R$. However, I do not see any elementary way to show that $g$ is $C^\infty$-smooth. (One could chop up the  Fourier transform of $g$ and thus reduce the problem to analytic functions, as one does in the proof of the Malgrange preparation theorem. But this looks like an overkill.)
Related posts: 


*

*The quotient of two functions  

*How to show a function is a test function?  

*Zeros of $C^\infty$ functions (which is the special case of the above, with accepted  answer that does not contain a   proof).    

 A: Applying Taylor's expansion with integral form of the remainder to $f$ at $0$, and noting that $f^{(k)}(0)=0$ for $k=0,\dots,n-1$, we have:
$$f(x)=\frac{1}{(n-1)!}\int_0^x(x-t)^{n-1} f^{(n)}(t)dt,\quad \forall x\in\Bbb R.\tag{1}$$
Substituting $t=sx$ into $(1)$, it follows that
$$f(x)=\frac{x^n}{(n-1)!}\int_0^1(1-s)^{n-1} f^{(n)}(sx)ds,\quad \forall x\in\Bbb R.\tag{2}$$
As a result, $g$ can be expressed as follows:
$$g(x)=\int_0^1h(s,x)ds,\quad \forall x\in\Bbb R, \tag{3}$$
where
$$h(s,x)=\frac{1}{(n-1)!}(1-s)^{n-1} f^{(n)}(sx),\quad  \forall (s,x)\in\Bbb R^2.$$
Evidently $h$ is a $C^\infty$ function on $\Bbb R^2$, so we can interchange the order of $k$-th differentiation w.r.t. $x$ and integration w.r.t. $s$ in $(3)$ freely for every $k\ge 1$. That is to say, $g$ is a $C^\infty$ function. 

Remark:


*

*Equation $(1)$ can be easily proved by using integration by parts repeatedly. 

*For the validity of differentiation under the integral sign, one may refer to this page.

*A slightly different approach to the problem is using induction on $n$ to reduce the problem to the $n=1$ case. More precisely, by induction on $n$, it suffices to prove that  $f_1(x)=\frac{f(x)}{x}$ can extend to a $C^\infty$ function on $\Bbb R$ and $f_1^{(k)}(0)=0$, $k=0,\dots, n-2$. The proof of smoothness of $f_1$ is similar, i.e. using the expression $f_1(x)=\int_0^1f'(sx)ds$. $f_1^{(k)}(0)=0~(0\le k\le n-2)$ follows from  applying Leibniz rule to $f(x)=xf_1(x)$ and induction on $k$. 

