O-notation: composing functions Big-oh and little-oh notation make things much simpler, and there are convenient rules for combining functions, for example, the ones mentioned here.
One rule conspicuously missing from the above list is a rule for function composition. If $f = O(\mu(x))$ at $a$, and $g=O(\nu(y))$ at $f(a)$, then what do we know about $g \circ f$ at $a$?
Some things which lead me to believe such a rule might exist, under certain conditions:


*

*Suppose $f$ and $g$ are $n$ times differentiable functions. It is a
theorem in calculus that if $p$ is the Taylor polynomial of degree
$n$ for $f$ at $a$, and $q$ is the Taylor polynomial of degree $n$
for $g$ at $f(a)$. Then $q \circ p$, truncated at degree $n$, is the
Taylor polynomial for $g \circ f$ at $a$. In other words, if  $$f= p
 - o((x-a)^n)\quad \text{and} \quad g = q + o((y-f(a))^n),\\ \text{then}\quad g \circ f = q \circ p + o((x-a)^n).$$

*In the simple case of polynomials, there is a very convenient
composition rule: Suppose $p(x)-p(a)=O((x-a)^n)$ at $a$, and $q(y) =
   O((y-p(a))^m)$ at $p(a)$. Then $q \circ p$ is $O((x-a)^{mn})$ at $a$.

 A: By the assumptions made, 
$$\limsup_{x\to a} \frac{|f(x)|}{|\mu(x)|}<\infty\quad \text{ and }\quad 
\limsup_{x\to a} \frac{|g(f(x))|}{|\nu(f(x))|}<\infty \tag1$$
If the function $\nu$ is doubling, meaning that there is $C$ such that 
$$ |s|\le 2|t|\implies |\nu(s)|\le C|\nu(t)| \tag2$$
then we can conclude from the first part of (1) that 
$$\limsup_{x\to a} \frac{|\nu(f(x))|}{|\nu(\mu(x))|}<\infty$$
and combining this with the second part of (1), 
$$\limsup_{x\to a} \frac{|g(f(x))|}{|\nu(\mu(x))|}<\infty \tag3$$
which is what one might have expected: $g\circ f=O(\nu\circ \mu)$.
Non-doubling case
Without (2) we don't get an explicit asymptotic like (3) because the constant implicit in the first part of (1) actually matters. For example, if $f(x)=O(x)$ as $x\to 0$ and  $g(x)=O(\exp(-1/x^2))$ as $x\to 0$, the asymptotic behavior of $g\circ f$ depends quite a bit on the implicit constant in $f(x)=O(x)$. We can state that 
$$g(x) = O(\exp(-C/x^2)) \quad \text{for some } \ C>0$$
which is less precise that having $O(\cdot )$ with a concrete comparison function in it.
Remark
Constant $2$ in the doubling condition (2) could be replaced by any constant greater than $1$; the meaning of the condition does not change. Doubling condition includes all polynomials (as in your example), but does not  include the functions growing or decaying at super-polynomial rate.
A: The argument about the Taylor polynomial is sound...but not every function's Taylor polynomial converges to the function, so it may have no impact on the functions themselves.  For instance, for 
$$
f(x) = \begin{cases} 1 - e^{-\frac{1}{x^2}} & x \ne 0 \\ 0 & x = 0\end{cases}
$$
the Taylor series converges everywhere (it's the zero series!), but converges to $f$ only at $x = 0$. So knowing something about the degree of the Taylor series doesn't tell you much about $f$. 
For nice enough functions, with series representations that converge everywhere, there might be something reasonable to say...but if $f$ and $g$ have everywhere convergent series, is it obvious that $f\circ g$ does as well? (It's not obvious to me!)
