I'm a little bit unsure if I use the Big-O-notation in the following context correctly:
Consider a function $\varphi \in C^{\infty}(\overline M)$ on a compact manifold with boundary and a boundary defining function $x$, so we have $\partial M = \{ x=0\}$ and $x>0$ on the interior $M$. Assume that we can use coordinates $(x,y)$ to describe points on $\overline M$.
Now we're interested in the behaviour of $\varphi(x)$ for $x \rightarrow 0$.
Assume that $\varphi \in o(x)$ for $x \rightarrow 0$. By definition this means, that $\lim \limits_{x\rightarrow 0} |\frac{\varphi(x)}{x}| = 0$. This means, that if we write $\varphi(x) = x \cdot g(x)$ that $g$ still is zero when $x$ is zero. Is this correct so far?
If this is correct, then one could write (could one, asymptotically?) $g(x) = x \cdot h(x)$ where $h(0)$ may is zero but we don't know.
The crucial thing for me now is the following:
Can one then say that $\varphi \in O(x^2)$ as $x \rightarrow 0$ because
\begin{equation} \lim\limits_{x\rightarrow 0} |\frac{\varphi(x)}{x^2}| = |h(x)| \Rightarrow \lim\sup\limits_{x\rightarrow 0} |\frac{\varphi(x)}{x^2}| = |h(x)| < \infty \end{equation} And, in general: Is it correct to treat a smooth function $\varphi$ that satisfies $\varphi \in O(x^k)$ asymptotically like $x^k\cdot g(x)$ for another smooth function $g$? (for $x \rightarrow 0$)
Thanks for any help!
Addendum:
I may formulated this issue a little bit confusing. The main question above is: Does $\varphi \in o(x)$ imply that $\varphi \in O(x^2)$ when $x \rightarrow 0$?