Taylor's Theorem says that for a function $f : \mathbb{R} \to \mathbb{R}$ which is $n$ times differentiable around a point $a$, there exists a function $h_n(x)$ such that $$f(x) = \left(f(a) + \sum_{k=1}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\right) + h_n(x)(x-a)^n$$ and $\lim_{x\to a} h_n(x) = 0$. I understand the theorem, its motivation and meaning and its proof. A quadratic approximation around a point $a$ exists then if we can write $f(x) = f(a) + b(x-a) + c(x-a)^2 + h_2(x)(x-a)^2$ where $h_2(x) \to 0$ as $x\to a$.
Now, looking deeper into this subject, I've noticed that the existence of a linear approximation implies the differentiability at the point so that they are equivalent (the other direction being proved via Taylor Theorem or otherwise). Indeed
$$ \frac{d}{dx}\big(h_1(x)(x-a)\big)|_{x=a} = \lim_{x\to a} \frac{h_1(x)(x-a)}{x-a} = \lim_{x\to a} h_1(x) = 0$$
So I've wondered whether the existence of such a quadratic approximation implies the existence of the second derivative at the point $a$. At first it seemed true to me but then I found this example:
Let $g : \mathbb{R} \to \mathbb{R}$, $g(x) = \begin{cases} 0, & x\notin\mathbb{Q} \\ (x-1)^3, & x\in\mathbb{Q} \end{cases}$. It's not hard to see that this function is only continuous at $x=1$ and the derivative exists there, $g'(1) = 0$. Then, if we define $u : \mathbb{R} \to \mathbb{R}, u(x) = \frac{g(x)}{(x-1)^2}$ and $u(1) := 0$, we have $u(x) \to 0$ as $x\to 1$.
So, for the function $f : \mathbb{R} \to \mathbb{R}, f(x) = 5 + 2(x-1) + 3(x-1)^2 + u(x)(x-1)^2$, the quadratic approximation around $x=1$ exists, but, since $f$ is only differentiable at that point, not around it, the second derivative there doesn't even make sense.
Is there anything wrong with this or the two concepts (the existence of the second derivative at a point and the existence of a best quadratic approximation around that point) are really not equivalent?