If $f:\mathbb{R}\rightarrow\mathbb{R}$, $f\in C^{\infty}(\mathbb{R})$ and $f(0)=0$ then $\frac{f(x)}{x}\in C^{\infty}(\mathbb{R})$ If $f:\mathbb{R}\rightarrow\mathbb{R}$, $f\in C^{\infty}(\mathbb{R})$ and $f(0)=0$ then $\frac{f(x)}{x}\in C^{\infty}(\mathbb{R})$.
Following ther is what i did:
the Maclaurin series for $f(x)$ is
$$f(x)=f(0)+\sum\limits_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!}x^n\stackrel{f(0)=0}{=}\sum\limits_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$
whence, we suppose $x\neq 0$,
$$g(x)\stackrel{\text{def}}{=}\frac{f(x)}{x}=\sum\limits_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n-1}$$
then $g(x)\in C^{\infty}(\mathbb{R}\setminus\{0\})$.
We can prove that
$$\lim_{x\rightarrow 0}g^{(k)}(x)=\frac{f^{(k+1)}(0)}{k+1}=g^{(k)}(0)\quad\forall k\in \mathbb{N}$$
Then $g(x)\in C^{\infty}(\mathbb{R})$.
Is it right?

We can write $f(x) = \int_0^x f'(t)\,dt = x\int_0^1 f'(xs)\,ds.$ So
$$g(x) = \int_0^1f'(xs)\,ds.$$
for all $x.$ You can differentiate all day through the integral sign, showing $g\in C^\infty.$
