<s>[INST] 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! :)
 [/INST] Always remember that $\mathcal{O}(g)$ is a set. So it is more appropriate to say that when $x\to a, f\in \mathcal{O}(g)$ which means that there exists an $\epsilon\gt 0 $ for which you can find a $\delta\gt 0$ such that $0\le |f(x)|\le \epsilon |g(x)|$ for all $0\lt|x-a|\lt \delta$. 
Your question 1): Why $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$ 
Answer: 
Note that $\lim_{x\to 0}\Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|=\frac 16$ hence by limit definition there exists a $\delta\gt 0$ such that 
$\begin{align}
0\lt|x|\lt \delta& \implies \Big|\Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|-\frac 16\Big|\lt 1\\
&\implies \Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|\lt \frac 76\\
& \implies \Big|\frac{x^3}{3!}+\frac{x^4}{4!}+...\Big|\le \frac 76 |x^3|\\
&\implies \frac{x^3}{3!}+\frac{x^4}{4!}+...\in \mathcal{O}(x^3) \end{align} $ 
Similarly, you can prove the second one.
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