If $f(x)$ is smooth and odd, must $f(x)/x$ be smooth?

Let $f:\mathbb R\to\mathbb R$ be:

1. smooth, i.e., infinitely differentiable, and
2. odd, i.e., $f(x)=-f(-x)$ for all $x$.

Let $g:\mathbb R\to\mathbb R$ be defined as

$$g(x):=\left\{ \matrix{f(x)/x, & x\neq 0 \\ \lim_{x\to 0} f(x)/x, & x=0} \right.$$

Must $g$ be smooth? In fact, is $g$ necessarily well defined at $x=0$?

This is not a HW problem, by the way. I am have been working on a different problem and just realized I have been assuming the above fact without proof. I am aware of some theorems in complex analysis that make this true for complex-analytic functions, but I also know that real analysis isn't always so clean.

$g$ will always be well defined at $0$ because:

$$g(0) = \lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \frac{f(x)-0}{x-0} \equiv f'(0)$$

by the definition of the derivative. One can even prove via repeated use of quotient rule and induction that by this definition,

$$g^{(n)}(0) = \frac{f^{(n+1)}(0)}{(n+1)}$$

which is well defined because $f$ is smooth.

Assume that $f:\mathbb{R}\to\mathbb{R}$ is smooth and $f(0) = 0$. If we define $g$ as in OP's construction, then $g$ is well-defined and the following representation holds for all $x \in \mathbb{R}$.

$$ g(x) = \int_{0}^{1} f'(xt) \, \mathrm{d}t. $$

Then it is easy to prove that Leibniz integral rule is applicable here, showing that $g$ is indefinitely differentiable and

$$ g^{(n)}(x) = \int_{0}^{1} \frac{\partial^n}{\partial x^n} f'(xt) \, \mathrm{d}t = \int_{0}^{1} t^n f^{(n+1)}(xt) \, \mathrm{d}t. $$

In particular, we immediately find that $g^{(n)}(0) = \int_{0}^{1} t^n f^{(n+1)}(0) \, \mathrm{d}t = \frac{f^{(n+1)}(0)}{n+1}$.