L'Hopital's Rule Complication $\textrm{Let } f(x) = \begin{cases} \displaystyle\frac{g(x)}{x}~, & \!\! x \neq 0 \\ 0~, & \!\! x = 0 \end{cases} \textrm{ for all } x \in \mathbb{R}.$
$\textrm{Assume } g(0) = g'(0) = 0 \wedge g''(0) = 17.$
$\textrm{Want To Prove } f'(0) = \displaystyle\frac{17}{2}.$
Information Necessary For The Use Of L'Hopital's Rule:
Important Lemmas
I showed that $f$ is continuous at $0$ so that $f'(0)$ can be computed using the limit definition.
In order to compute $f'(0)$, I wish to use L'Hopital's Rule (LHR).
$f'(0)$
$= \displaystyle\lim_{x \rightarrow 0} \frac{f(0 + x) - f(0)}{x}$
$= \displaystyle\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$
$= \displaystyle\lim_{x \rightarrow 0} \frac{f(x) - 0}{x}$
$= \displaystyle\lim_{x \rightarrow 0} \frac{f(x)}{x}$
$= \displaystyle\lim_{x \rightarrow 0} \frac{\displaystyle\frac{g(x)}{x}}{x}$
$= \displaystyle\lim_{x \rightarrow 0} \frac{g(x)}{x^2}$
$= \displaystyle\lim_{x \rightarrow 0} \frac{g'(x)}{2x}~~~$ According to LHR: Justification
In order to get $\displaystyle\frac{17}{2}$, I must use L'Hopital's Rule again to get $\displaystyle\lim_{x \rightarrow 0} \frac{g''(x)}{2}$.
However, so far, I cannot show that $g''(x)$ is never $0$ as $x \rightarrow 0$, which is an important property justifying the use of LHR.
In other words, I have no continuity assumption or third derivative to prove there exists an interval containing $0$ where $g''(x)$ is never $0$, using the information I have derived.
This is my attempt so far of justifying my second use of LHR:
Justification 2
As you can see, I am missing the final step regarding $g''(x)$.
Any help would be appreciated.
Let me know if there is any function $g$ where $f'(0)$ wouldn't be $\displaystyle\frac{17}{2}$, in which case I must assume something about $g$ or $g''$.
 A: Note that we need not appeal to L'Hospital's Rule.  If $f'(0)$ exists, then it is equal to the limit given by
$$\begin{align}
\lim_{x\to 0}f'(0)&=\lim_{x\to 0}\frac{\frac{g(x)-g(0)}x}{x}\\\\
&=\frac12\lim_{x\to 0}\frac{g(x)-2g(0)+g(-x)}{x^2}\\\\
&=\frac12g''(0)
\end{align}$$
And we are done!

Alternatively, we may apply L'Hospital's Rule once and write
$$\begin{align}
\lim_{x\to 0}f'(0)&=\lim_{x\to 0}\frac{\frac{g(x)-g(0)}x}{x}\\\\
&\overbrace{=}^{\text{LHR}}\lim_{x\to 0}\frac{g'(x)}{2x}\\\\
&=\frac12\lim_{x\to 0}\frac{g'(x)-g'(0)}{x}\\\\
&=\frac12g''(0)
\end{align}$$
A: if $\phi(x) \to 0$ as $x \to 0$, and also $\psi(x) \to 0$ as $x \to 0$, and if you are interested in
$$\lim_{x \to 0}\,\frac{\phi(x)}{\psi(x)},$$
and if also $\phi$ and $\psi$ are differentiable at $0$, then
$$\lim_{x \to 0}\,\frac{\phi(x)}{\psi(x)}\,=\,\frac{\phi'(0)}{\psi'(0)},$$
if it also is true that $\psi'(0)$ is not zero. This lemma is enough for your problem.
You want to show:
$$f'(0)\,=\,\lim_{h \to 0}\,\frac{f(h)}{h}\,=\,\lim_{h \to 0}\,\frac{g(h)}{h^2}\,=\,\frac{17}{2}.$$
By L'hopital, since $g$ is continuous at zero,
$$\lim_{h \to 0}\,\frac{g(h)}{h^2}\,=\,\lim_{h \to 0}\,\frac{g'(h)}{2h},$$
if the latter exists. We have again a "zero/zero" form, since $g'$ is continuous at zero. But $g'$ is in fact differentiable at zero, so we use the lemma:
$$\,\lim_{h \to 0}\,\frac{g'(h)}{2h}\,=\,\frac{g''(0)}{2}.$$
(Alternatively -- as pointed out by Mark Viola -- this last result follows directly from the definition of $g''(0)$.)
