small o and big O Okay, so I need some help grasping the "big O, small o" concept. I'd put up what I think I've understood, and then maybe you can correct my mistakes and enlighten me further. 
small o
If we let $x$ approach some $a$, then if $f(x)/g(x)$ approaches 0, we say $f(x) = o(g(x)$. Now, if $a = 0$, then this means what exactly? That $f(x)$ approaches $0$ faster than $g(x)$? But what if $a \neq 0$? Then how do we give meaning to the fraction going to zero? Why would that happen?
big o
So if $f(x) = O(g(x))$, then $$||f(x) || \le K||g(x)||$$ for some K (even though it only is true if we are near enough the limit $a$). But 1) how do you actually show the above? Most examples I see involve some polynomial of nth degree, but that's an easy example. What if we're working with sines, cosines, logs, exponentials, etc, and not just simple polynomials? 2) What does it actually mean, intuitively, to be big O of something, and how does it relate to being small o? If $f$ is big O of $g$, is it also small o? Or vice-versa?
 A: Your understanding of the small-$o$ is ok. If $a\neq 0$, then $a\neq 0$ and there is no problem with that. To say that $f(x) = o(g(x))$ as $x\to a$ means that $f$ gets infinitely smaller than $g$ as $x$ approaches $a$, or more precisely, $\lim_{x\to a} \left|f(x)/g(x)\right|=0$. This can happen in lots of different ways. For example, it may be the case that $f$ approaches $0$, but $g$ doesn't. Or $g$ approaches $\infty$ but $f$ doesn't. Or $f$ and $g$ both approach $0$, but $f$ does so muich faster than $g$.
As to the big-$O$, the intuitive meaning is that, when you say "$f(x)=O(g(x))$ as $x\to a$", you mean that "$f$ doesn't grow infinitely larger than $g$ as $x\to a$". Again, this can be made clearer by writing $\limsup_{x\to a} \left|f(x)/g(x)\right| < \infty$. Your equation $|f|\leq K|g|$ can be justified by defining $K$, for example, as $K=\limsup_{x\to a} \left|f(x)/g(x)\right|$. Then, as $x$ gets closer to $a$, there will be a moment when you can write $|f|\leq (K+\varepsilon)|g|$. If you want an easier, strict inequality, you can say that, if $K$ is the limsup mentioned above, then $|f|\leq (K+1)|g|$ for $x$ close enough to $a$.
You also asked how to prove it when $f$ is big-$O$ of $g$. Well, one way is to show that, for $x$ in some interval around $a$, $f/g$ is bounded (since $|f|\leq K|g|$ is the same as $|f/g|\leq K$). One other way is to calculate the limit of $f/g$, if it converges (since the limit is equal to the $\limsup$ when the limit exists). If nothing else works, you can show directly that $\limsup_{x\to a} |f/g|$ is a real number (less than infinity).
I hope that the characterizations "$f$ gets infinitely smaller than $g$" and "$f$ doesn't grow infinitely larger than $g$" given above are sufficient to show that, if $f(x)=o(g(x))$ then $f(x)=O(g(x))$, but not the other way around.
A: Using fraction in definitions reduce generality, because of possibility of $a$ to be limit point for $g$'s zeros.
Taking for simplicity  case  $f:\mathbb{N}\longrightarrow\mathbb{R}_{\geq 0}$ and $g:\mathbb{N}\longrightarrow\mathbb{R}_{\geq 0}$ we have following definitions:
$$O(g) = \left\lbrace f:\exists C > 0, \exists N \in \mathbb{N}, \forall n (n >  N \& n \in \mathbb{N}) (f(n) \leqslant C \cdot g(n)) \right\rbrace$$
$$o(g) = \left\lbrace f:\exists \epsilon(n) \geqslant 0, \lim_{n \to \infty}\epsilon(n)=0, \exists N \in \mathbb{N}, \forall n (n >  N \& n \in \mathbb{N}) (f(n) = \epsilon (n) \cdot g(n)) \right\rbrace$$
So, both are sets and first gives some "intellectual" boundary for its $f$ elements, while second gives some kind of representation for its $f$ elements in $a$'s neighbourhood. Main relationship:
$$o(g) \subset O(g)$$
