Local definition and properties of higher derivatives The functions considered are real and defined in a neighborhood of $0$.
Observation: $f(0) = 0$ and $f(x)$ is differentiable once at $x = 0$ if and only if $f(x) / x$ is continuous at $x = 0$.
Can this idea be generalized to higher derivatives?
If $g(0) = g'(0) = 0$ and $g(x)$ is differentiable twice at $x = 0$, is then $g(x) / x^2$ continuous at $x = 0$ ? It is true with the additional hypothesis that $g''(x)$ is continuous, but is it a necessary condition ?
If $h(x)$ is continuous at $x = 0$, is then $x^2 \cdot h(x)$ differentiable twice at $x = 0$ ?
The difficulty is that it is not known whether the functions considered are continuous resp. differentiable unless $x = 0$.
 A: I finally managed to solve this problem about the subtleties of the definition
of higher derivatives.
Assuming that $g(x)$ and $g'(x)$ exist and are continuous in a neighborhood of
$x = 0$, that $g(0) = g'(0) = 0$, and that $g''(0)$ exist, we have:$$
\begin{align}
g''(0) &= \lim_{ x \to 0 } \,(g'(x) - g'(0)) \mathbin{/} (x - 0)
       & \textrm{(definition)} \\
       &= \lim_{ x \to 0 } g'(x) \mathbin{/} x
       & \textrm{($g'(0) = 0$)} \\
       &= \lim_{ x \to 0 } 2 \cdot g(x) \mathbin{/} x^2
       & \textrm{(L'Hospital's rule)} \\
\end{align}
$$
That is, $u(x) := 2 \cdot g(x) \mathbin{/} x^2$ is continuous at $x = 0$,
$g(x) = x^2 \cdot u(x) \mathbin{/} 2$, and $g''(0) = u(0)$.
Remark: L'Hospital's rule establishes the existence of
$\lim_{ x \to 0 } 2 \cdot g(x) \mathbin{/} x^2$
from the existence of
$\lim_{ x \to 0 } g'(x) \mathbin{/} x$.
Remark: It is not required that g''(x) exists, except at $x = 0$. In
particular, $\lim_{ x \to 0 } g'(x) \mathbin{/} x$ is not necessarily equal to
$\lim_{ x \to 0 } g''(x)$, because the latter may not exist (even if $g''(x)$
exists in a neighborhood of $x = 0$, it may not be continuous at $x = 0$).
Remark: This can be generalized to higher derivatives. For instance:$$
\begin{align} 
g'''(0) &= \lim_{ x \to 0 } \,(g''(x) - g''(0)) \mathbin{/} (x - 0) \\
        &= \lim_{ x \to 0 } g''(x) \mathbin{/} x                    \\
        &= \lim_{ x \to 0 } 3! \cdot g(x) \mathbin{/} x^3           \\
\end{align}
$$
As to $v(x) = x^2 \cdot h(x)$ with $h(x)$ continuous at $x = 0$, it is not
differentiable twice, because $h(x)$ may be discontinuous everywhere except
at $x = 0$, such that $v'(x)$ may not exist anywhere except at $x = 0$ (where
it exists, namely $v'(0) = 0$). However, the definition of the second
derivative could be consistently extended such that $v''(0) = 2 \cdot h(0)$.
