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! :)
 A: 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 $|x| < \frac{1}{2}$.
A: When we write $f(x)=O(x^n)$ as $x\to 0$, what this means is that there exist $M>0$ and $\delta>0$ such that for all $|x|<\delta$, we have $\left|f(x)\right|\leq M|x|^n$. Equivalently, there must exist a $\delta>0$ such that
\begin{align}
\sup_{0<|x|<\delta}\left|\frac{f(x)}{x^n}\right|<\infty.
\end{align}
In plain English: the function divided by $x^n$ must be bounded in some punctured neighborhood of the origin. This is just definition.

*

*We have $f(x):= \sum_{k=3}^{\infty}\frac{x^k}{k!}=x^3\left(\sum_{k=3}^{\infty}\frac{x^{k-3}}{k!}\right)$. In this case we are very fortunate because ANY $\delta>0$ works, and there is no need to show the existence of a specific one. This is because if $|x|\leq \delta$ then
\begin{align}
|f(x)|\leq |x^3|\cdot \left(\sum_{k=3}^{\infty}\frac{\delta^{k-3}}{k!}\right)
\end{align}
If you want to pattern-match with the definition, then the bracketed term here is $M$, and this is clearly finite because the entire series defining the exponential converges (absolutely) at every point (heck if one wants a very crude estimate then we can bound the series by $\frac{e^{\delta}}{\delta^3}$, which is clearly finite).

As for 2, I think you should atleast write down your attempt at a proof based on your understanding of the definitions.
A: 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.
A: For $0\le x\le1$,

*

*$\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\cdots\le x^3\left(\dfrac1{3!}+\dfrac1{4!}+\dfrac1{5!}+\cdots\right)$ and the sum inside the parenthesis is finite.


*Similarly, $\dfrac{x^2}2+\mathcal O(x^3)\le \dfrac{x^2}2+cx^3\le x^2\left(\dfrac12+c\right).$
