My calculus book states that:
We write $f(x)=\mathcal{O}(u(x))$ as $x\to a$ provided that $|f(x)|\le K|u(x)|$
This means that for example as $x \to 0$:
$$e^x-P_1(x)=\mathcal{O}(x^2)$$
...where $P_1(x)$ is the first order Taylor Polynomial for $e^x$.
What I do not understand is what the use of this statement is when talking about Taylor polynomials.
For example: aren't the following statements also true as $x \to 0$?
$$e^x-P_1(x)=\mathcal{O}(x+1)$$
$$e^x-P_1(x)=\mathcal{O}(1)$$
$$e^x-P_1(x)=\mathcal{O}(x^3+1)$$
For all of these examples we can find some $K$ for which the statement is valid close to $0$, isn't it?
I either don't understand something about Big-Oh notation, or I don't understand the use of Big-Oh notation in Taylor Polynomials. What is so special about $e^x-P_1(x)=\mathcal{O}(x^2)$?
Please help me to shed some light on this issue.