Assume we are given smooth functions $f, g: U \to \mathbb{C}$, where $0 \in U \subset \mathbb{R}^n$ is open and $0 \in g^{-1}(0) \subset \{x_n = 0\}$. Furthermore, suppose that $\nabla g \neq 0$ on the set $g^{-1}(0)$ and let $h = \frac{f}{g}$ defined on $U \setminus g^{-1}(0)$. Assume that all the partial derivatives $D^\alpha h$ for $k \geq 0$ are bounded in $\{x_n > 0\}$. Does there exist a smooth extension $\tilde{h}$ of $h$ to the whole of $U$?
Here's something that I've tried (my intuition is that such an extension exists):
Because all the derivatives are bounded in $x_n > 0$ we easily conclude that $h$ is uniformly continuous and hence there exists a unique smooth continuation to $\{x_n \geq 0\}$. However, I wasn't able to relate this to the {x_n < 0} region.
For n = 1, this is basically l'Hospital's rule and such an extension clearly exists. We can use this on the set of lines given by varying $x_n$ and keeping $(x_1, \dotso, x_{n - 1})$ constant (and assuming $\frac{\partial g}{\partial x_n} \neq 0$).
If there's an open set $V \subset g^{-1}(0)$, then we may use Taylor's theorem to conclude that $g = x_n g_1$ with $g_1 \neq 0$ near $V$. Since $h$ is bounded, we must have $f = x_n f_1$ near $V$ and so $h$ extends. How do we then extend over the remaining, nowhere dense set?
