Continuous extension of $f(x)/x$

Question

I have the problem:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a smooth function such that $f(0)=0$ and consider $ F $ the continuous extension of $ \frac{f (x)}{x} $, is $ F $ a smooth function?

I suppose the result is true, and I know that:

$f'(0) = \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x} = \lim_{x \rightarrow 0} \frac{f(x)}{x} = F(0)$

$f''(0) = \lim_{x \rightarrow 0} \frac{ f'(x) - f'(0)}{x} = \lim_{x \rightarrow 0} \frac{ f'(x) - F(0)}{x} $

I want to use a reasoning like this to define $ F^{(n)} (0) $, does anyone have a counterexample or demonstration?

Answer 1

(This answer is taken from a french math forum.)

We have $f(x)=x\int_0^1f'(xt)dt$, therefore for all $x\neq 0$, $\frac{f(x)}{x}=\int_0^1f'(xt)dt$. By continuity-integral, the right hand side is a continuous function on $x$, therefore $F(x)=\int_0^1f'(xt)dt$. By smoothness-integral, $F$ is a smooth function.