Big O notation and polynomials I often see that the following polynomial can be written as such:
$f(x) = 6x^4+3x^3+O(x^2)$
where the big O collects all the lower order terms.
Yet, I also see this sometimes:
$f(x) = 6x^4+7x^9+O(x^{11})$
where the big O collects higher order terms.
How does this even make sense? For $f(x)$ to be in $O(g(x))$ doesn't it need to grow at least as fast as $g(x)$ at super large values of $x$?
 A: To expand on what  Bernard said:
In general,
big-Oh and little-Oh terms
are smaller than
the explicit terms
that they are together with.
When you have
$f(x) = 6x^4+3x^3+O(x^2)$,
this intends
$x^2$
to be smaller than
$x^3$ and $x^4$.
This holds when
$x$ is large,
so when
$x \to \infty$.
When you have
$f(x) = 6x^4+7x^9+O(x^{11})
$,
this intends
$x^{11}$
to be smaller than
$x^9$ and $x^4$.
This holds when
$x$ is small,
so when
$x \to 0$.
A: The Big-O terms in the function definitions are place-holders. They are just saying that there is a function that is asymptotically bounded above by $x^{11}$. The $O(x^{11})$ isn't collecting higher order terms. A function $g(x)$ containing $x^{12}$ for example, wouldn't be a part of $f(x)$.
The Big-O notation is commonly used as short-hand for this purpose.
A: You should write something like:
$f(x) \in 6x^4+3x^3+O(x^2)$ as $x \to \infty$
$f(x) \in 6x^4+3x^3+O(x^{11})$ as $x \to 0$
There are two points here. Firstly the Big-O notations denote classes of functions, not a single function, and it is a misleading abuse of notation to use "$=$" between a single function and a class. Don't forget that $O(x) - O(x) = O(x)$, not $O(x) - O(x) = 0$. Secondly, we cannot talk about limiting behaviour without specifying the region we are concerned with, just like it is meaningless to write "$\frac{2^n}{n^2} \to \infty$" just like that without specifying anything about $n$.
