I tried to prove the following lemma:
Lemma. Let $f,g:\mathscr U_{x_0}\to \mathbb R$ differentiable $n$ times in $x_0$. Then $$f(x)-g(x)=o((x-x_0)^n)\iff f^{(k)}(x_0)=g^{(k)}(x_0),\ 0\le k\le n$$
Proof. (by induction)
The claim is true for $k=0$, let's suppose it is also true for $k=n$, and let's prove it for $k=n+1$:
"$(f(x)-g(x)=o((x-x_0)^{n+1})\implies f^{(n+1)}(x_0)=g^{(n+1)}(x_0))$"
$$\lim_{x\to x_0}\frac{f(x)-g(x)}{(x-x_0)^{n+1}}= 0$$
Now my thought is using L'Hopital $n$ times to achieve
$$\boxed{\lim_{x\to x_0}\frac{f(x)-g(x)}{(x-x_0)^{n+1}}= \frac{1}{(n+1)!}\lim_{x\to x_0}\frac{f^{(n)}(x)-g^{(n)}(x)-\overbrace{(f^{(n)}(x_0)-g^{(n)}(x_0))}^{= 0\text{ for induction }}}{x-x_0}=0}$$
"$(f^{(n+1)}(x_0)=g^{(n+1)}(x_0)\implies f(x)-g(x)=o((x-x_0)^{n+1}))$"
$$\lim_{x\to x_0}\frac{f^{(n)}(x)-g^{(n)}(x)}{x-x_0}=0$$ Applying L'Hopital the other way around (that is, going back to the primitive), I got
$$\boxed{\lim_{x\to x_0}\frac{f^{(n)}(x)-g^{(n)}(x)}{x-x_0}=n!\lim_{x\to x_0}\frac{f(x)-g(x)}{(x-x_0)^{n+1}}=0}$$
I am not sure about the boxed equalities.