If $f, g$ are smooth functions on $I$, $g$ has no flat point, and $f/g$ has a continuous extension to $I$, is this extension smooth?

Question

Let $f,g$ be smooth functions $I \to \mathbb{R}$. Suppose that the following two conditions are met:

1. For all $t \in I$, there exists $m \in \mathbb{N}$ such that the $m$-th derivative $g^{(m)}$ is nonzero at $t$. (This implies, in particular, that every zero of $g$ is isolated.)
2. The function $f/g$ admits a continuous extension to $I$. (This happens when the order of every zero of $g$ is smaller than or equal to the one of $f$.)

Does it then follow that such extension is smooth?

Answer 1

Yes.

Claim: If $g(x_0)=0$, then there is a smooth $\tilde{g}\colon I\to\mathbb{R}$ satisfying $g(x)=(x-x_0)\tilde{g}(x)$ for all $x$.

Proof: The only point need checking is $x_0$. For any $n\in\mathbb{N}$, Taylor expand $g$ about $x_0$ up to $(x-x_0)^{n+1}$ with remainder term, and divide by $x-x_0$ to get Taylor expansion of $\tilde{g}$ up to degree $n$ with remainder. QED.

So $f(x)=(x-x_0)^m F(x)$, $g(x)=(x-x_0)^n G(x)$ near $x_0$, $F,G$ smooth and $m\geq n$. Cancel the $(x-x_0)$s, and you have product and quotient of smooth functions $(f/g)(x)=(x-x_0)^{m-n}F(x)/G(x)$ near $x_0$.