Proof of repeated applications of L'hopital rule at a point NOTES: I will not be assuming that the domains here after are an interval. For $f'(x)$ to be defined all I require is that the usual limit exists and $x$ is an accumulation point of the domain (may or may not be an interior point). Also I'll just be concentrating on $f(x)=0$ as $x \to 0$ for simplicity sake.
For a concrete example take the identity function restricted to $\mathbb{Q}$, $0$ is an accumulation point and you can see the limit is well defined. This definition is in Tao's Analysis books, for example. So $f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{(x+h)-(x)}{h}=1$. This holds because every $x$ in $\mathbb{Q}$ is an accumulation point of $\mathbb{Q}$.

Some terminology:
One-time L'hopital: If $\lim \limits_{x \to 0} f(x)=0$ and $\lim \limits_{x \to 0} g(x)=0$, and $f'(0)$ and $g'(0)\not = 0 $ exist then $\lim \limits_{x \to 0} \frac{f(x)}{g(x)}=\frac{f'(0)}{g'(0)}$.
Repeated L'hopital: If $\lim \limits_{x \to 0} f(x)=0$ and $\lim \limits_{x \to 0} g(x)=0$, and $f^{(a)}(0)=g^{(a)}(0)=0$  for $0<a<n$, and $f^{(n)}(0)$ and $g^{(n)}(0) \not = 0$ exist then $\lim \limits_{x \to 0} \frac{f(x)}{g(x)}=\frac{f^{(n)}(0)}{g^{(n)}(0)}$.
I've only seen proofs of the repeated L'hopital rule when the domain  of the function is an interval around $0$, while it's easy to see that the one-time L'hopital holds even when $x$ is not an interior point and just an accumulation point. My question is, is the repeated L'hopital true even if $x$ is only an accumulation point and not an interior point? Preferably without using Taylor's Theorem as I've found the same problem there, I've only seen proofs of it only when $x$ is an interior point.
 A: Without proper definitions it is hard to prove or disprove any statement.  So let us try to define the derivative of $f:S \to {\Bbb R}$ defined on the dense set $S\in[0,1]$. Let us say that the derivative at $x\in S$ is given by:
$$   f'(x)=\lim_{y\to x, y\in S} \frac{f(y)-f(x)}{y-x},$$
provided the limit exists. If is does for every $x\in S$ then we have obtained a function $f':S \to {\Bbb R}$ and can repeat the process to produce $f'':S\to {\Bbb R}$ etc... We do not try to extend $f'$ to a larger subset.
With this definition, your claim does not work for $n\geq 2$. An example: Let $S={\Bbb Q}\cap [0,1]$, $\alpha_0=1$ and let $\alpha_k$ be any strictly decreasing sequence of irrationals in $(0,1)$ going to zero and such that $\alpha_{k}/\alpha_{k+1}\to 1$ when $k\to +\infty$.
Define the intervals  $I_k=(\alpha_{k},\alpha_{k-1})_{k\in {\Bbb N}}$.
Then $\bigcup_{k\geq 1} I_k \supset S\cap (0,1)$.
For $x\in I_k\cap S$,
we set $f(x) = (\alpha_k+\alpha_{k-1})^2/4 $ and we let $f(0)=0$.
Then $\lim_{x\to 0^+} f(x)/x^2=1$. However,  every rational $x\in (0,1)\cap S$ is in the interior of some interval $I_k$ so $f$ is locally constant at $x$ and $f'(x)=0$ with our definition. Letting $x\to 0$ we also get $f'(0) = \lim_{x\to 0^+,x\in S}  f(x)/x=0$. Our derivative is thus identically 0 on $S$ and we also deduce that $f''(x)=0$ for every $x\in S$. But then:
$$\lim_{x\to 0^+,x\in S}  f(x)/x^2 =1 \neq 0 =\lim_{x\to 0^+, x\in S} f'(x)/2x= \lim_{x\to 0^+,x\in S} f''(x)/2$$
