Big O for error terms The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the  example give here  we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.
1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$
Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?
2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?
I appreciate your help! :)

The intuitive answer to why
$$\frac{x^3}{3!}+ \frac{x^4}{4!} + \cdots = \mathcal{O}(x^3)$$
as $x \to 0$ is that $x^3$ is the term that goes to $0$ at the lowest rate. All higher-order terms go to $0$ faster, so the expression will "at worst" approach $0$ like $x^3$. More precisely, the reason in that
$$|x^3| \geq C|x^4 + x^5 + \cdots |$$
for $x$ small enough. To see this, observe that for $|x|<\frac{1}{2}$, we have
$$\left| x^4 + x^5 + \cdots \right| = \left| \frac{x^4}{1-x} \right| = \left| \frac{x}{1-x} \right| |x^3| \leq C |x^3|$$
since $\frac{x}{1-x}$ is bounded for $|x| < \frac{1}{2}$.
