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Currently I am trying to understand landau big O notation with this post on the wikipedia

$$\exp\left(x\right)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdot\cdot\cdot~~~~~\text{for all}~~x\tag{1}$$

$$=1+x+\frac{x^2}{2}+\mathcal{O}\left(x^3\right)~~~~\text{as}~x\to 0\tag{2}$$

$$=1+x+\mathcal{O}\left(x^2\right)~~~~\text{as}~x\to 0\tag{3}$$

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I've interpreted the last statement above as follows.

$$\lim_{x\to 0}\left|\frac{\exp\left(x\right)-1-x-\frac{x^2}{2}}{x^3}\right|=M\tag{4}$$

$$~~\Leftrightarrow~~\lim_{x\to\infty}\frac{\left|\exp\left(x\right)-1-x-\frac{x^2}{2}\right|}{\left|x^3\right|}=M\tag{5}$$

$$~~\Leftrightarrow~~\lim_{x\to 0}\left|\exp\left(x\right)-1-x-\frac{x^2}{2}\right|=M\cdot\left|x^3\right|\tag{6}$$

Since,$~g\left(x\right)>0~$must always be satisfied, I can rewrite the eqn4 into the following one.

$$\lim_{x\to 0^{+}}\left|\frac{\exp\left(x\right)-1-x-\frac{x^{2}}{2}}{x^{3}}\right|=M~~\leftarrow~~\text{Approach to zero from positive side}\tag{7}$$

$$=\lim_{x\to 0^{+}}\left|\frac{\exp\left(x\right)-1-x-\frac{x^{2}}{2}}{x^{3}}\right|\tag{8}$$

$$=\lim_{x\to 0^{+}}\frac{\left|\exp\left(x\right)-1-x-\frac{x^2}{2}\right|}{\left|x^{3}\right|}\tag{9}$$

$$=\lim_{x\to 0^+}\frac{\left|\exp\left(x\right)-1-x-\frac{x^2}{2}\right|}{x^3}\tag{10}$$

About above, I can't determine whether absolute braces can be removed

As those absolute symbols are removed, then I know the following.

$$L_{\text{positive}}:=\lim_{x\to 0^{+}}\frac{\exp\left(x\right)-1-x-\frac{x^2}{2}}{x^{3}}=\frac{1}{6}\tag{11}$$

Why the wikipedia article seems asserted to remove the absolute symbols in eqn10?

How$~x^3~$of$~\mathcal{O}\left(x^3\right)~$is determined so quickly in eqn2?

electrical apprentice
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    Technically speaking $O(x^3)$ is a collection of functions, so one should write things like $f(x) \in O(x^3)$. However, people often make the abuse of notation $f(x) = O(x^3)$, or substitute $O(x^3)$ into expressions as if it were a function of $x$. That said, I don't understand why you would want to write $O(x^3) = f(x)$ as in your (6), with the $O(x^3)$ on the left-hand side. – angryavian Dec 06 '21 at 00:25
  • As$~x~$approaches to infinity, then I know that$~\mathcal{O}\left(x^{n}\right)=\mathcal{O}\left(x^{n+1}\right)~(n\in\mathbb{N})~$is true and$~\mathcal{O}\left(x^{n+1}\right)=\mathcal{O}\left(x^{n}\right)~$is wrong. So simply wondered the same thing(order of placements)with small$~x~$ – electrical apprentice Dec 06 '21 at 01:37
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    Indeed, the proper statement is $O(x^n) \subseteq O(x^{n+1})$. As I mentioned in my previous comment, usage of "$=$" is a common abuse of notation that can lead to misunderstandings like yours. – angryavian Dec 06 '21 at 02:00
  • One way: You can use L'Hopital's rule multiple times on $$\lim_{x\rightarrow 0} \frac{e^x - 1 - x - x^2/2}{x^3}$$ Another is, for $-1<x<1$, $$ \frac{|e^x - 1 - x - x^2/2|}{|x^3|} \leq \frac{1}{3!} +|x| \sum_{i=4}^{\infty}\frac{1}{i!}$$ – Michael Dec 06 '21 at 04:14
  • @Michael It converged to $~\frac{1}{6}$ – electrical apprentice Dec 06 '21 at 04:21
  • Yes that limit gives $\frac{1}{3!}=\frac{1}{6}$. – Michael Dec 06 '21 at 04:23
  • I've changed my this post drastically. – electrical apprentice Dec 08 '21 at 23:54

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