Is $f(x)/x$ differentiable at $x=0$? Let $\varepsilon> 0$, $I:=(-\varepsilon,\varepsilon)$ and $f\in C^\infty(I,\mathbb R)$ with $f(0)=0$. By the definition of the derivative we know that
$$h:I\setminus\{0\}\to \mathbb R, \qquad x \mapsto \frac{f(x)}{x}$$
can be continuously extended to $I$ with $h(0):=f'(0)$. But what do we know about differentiability of $h$ at $0$? Is $h\in C^\infty(I,\mathbb R)$?
EDIT: For analytic functions this is clearly the case. But what do we have in general?

Yes, you may write (the first for non-zero $x$):
$$ h(x) = \frac{1}{x}\int_0^x f'(t)\; dt = \int_0^1 f'(u x) \; du $$
from which the result follows as you may take derivatives inside the last integral.
