Sense of big-O notation in real analysis I was solving some problems with big-O notations. I know the definition or how to prove it, but I still can't "feel" their main usage exactly in real analysis (not in computer science while investigating complexity of algorithm)
Like what does these expressions even mean: $2x-x^2 = O(x), x \to 0$ or $x \sin \frac{1}{x} = O(|x|), x \to 0$? 
When we use little-o notation, $f = o(g), x \to x_0$, I understand that $f$ is infinitesimal of a higher order than $g$. But when it comes to $f = O(g), x \to x_0$, I can't somewhy see through definition its sense. I even plotted their graphs - still no clue
 A: You can imagine $o$ and $O$ as analogues of $>$ and $\geq$. Basically, $f\in o(g)$ means that $f$ is of higher order than $g$, while $f\in O(g)$ means that $f$ is at least of order $g$.
For instance, consider $x^r$ as $x\to0$. This is $o(x^s)$ for all $s<r$, but it is not $o(x^r)$.
At the same time, it is $O(x^s)$ for all $s\leq r$. In particular, $x^s$ is obviously of the same order as itself (so $O$ of itself), but not of higher order than itself (so not $o$ of itself).
A: A  slightly nonstandard but equivalent way to define the statement

$ f= O(g)$  as $x\to 0$

is to interpret $O$ as a wild-card symbol for some unspecified  function that is bounded near $x=0$; and then you  can interpret the expression $O(g)$ as the product  $ O \cdot g$  ( the product of that  bounded function and $g$.) This interpretation streamlines many formal manipulations involving such asymptotic relations, especially once you establish the "fundamental identities" $O \cdot O\subset O$ and  $o\cdot O \subset o$.
P.S> To make this notation more rigorous, you can regard $O$ as the collection of all bounded functions,  and then regard these fundamental identities as statements about any pair of elements chosen from these collections.
For example $ f(x)= 2x - x^2 = 2x (1- \frac{x}{2}) $ can be written as a bounded multiple of $ g(x)=2x$. The bounded expression $(1- \frac{x}{2})$ belongs to the collection denoted by $O$.
A: To say that $2x-x^2 = O(x)$ as $x\to 0$ is to say that $|2x-x^2|\le C|x|$ for some constant $C>0$, or, rephrasing, that $\left|\dfrac{2x-x^2}x\right|\le C$. Since for $|x|\le 1$ we have $x^2\le |x|$, we in fact have $|x|\le |2x-x^2|\le 2|x|$ for all $x$ with $|x|\le 1$. So we can take $C=2$.
In the case of your second example, since $|\sin u|\le 1$, we have $|x\sin\frac1x| \le |x|$, so here we can take $C=1$.
When we say $f(x)= O(g(x))$ as $x\to x_0$, we're saying that some multiple of $g(x)$ dominates $f(x)$ for $x$ near $x_0$.
