Understanding $f(x)=O(g(x))$ as $x \to c$ of two infinitely-small functions I'm trying to get an intuition behind the following definition:

Let $f(x)$ and $g(x)$ be infinitely-small functions when $x \to c$, i.e.
  $\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0$. We say that $f(x)$ is
  infinitely-small function with not lower order than $g(x)$ ($f$ converges
  to $0$ not slower than $g$) when $x \to c$, if there exists a constant $M 
> 0$ s.t. $|\frac{f(x)}{g(x)}| \le M$ for all $x$ sufficiently close to
  $c$.
We use the following notation: $f(x)=O(g(x))$.

Now, I don't really get what 'sufficiently close to $c$' means, how is this described symbolically? Also, how can I interpret this definition visually?
 A: Sufficiently close simply means there is some $\delta > 0$ such that the condition holds for $x \in (c - \delta, c + \delta) \setminus \{ c \}$.
To get a feel for what this means visually, look at the graphs of $x$ and $x^2$ around the origin. Clearly $x^2 < x$ sufficiently close to $0$, and so we say $x^2 = O(x)$. However, notice that $\frac{1}{2}x$ is also 'lower than' $x$, so it too is $O(x)$. The function $2x$, however, fits the definition too, and yet it is larger than $x$ near $0$. What's going on? The idea is that $2x$ isn't approaching $0$ faster in any 'fundamental way' than $x$ is; up to constants, they approach $0$ equally fast. The same is not true for $x^2$, however: no matter how large you scale up $x^2$, making it $C x^2$ for some really large $C$, you will always be able to zoom in enough to find that $Cx^2 < x$ around $0$. Hence it doesn't just approach $0$ 'not slower' than $x$, but faster even. This is noted by a lowercase $o$: $x^2 = o(x)$. Rigorously, the small $o$ simply requires that the limit in the defintion of $O$ be zero, not just finite.
A: Roughly speaking, "sufficiently close to $c$" means in some (punctured) neighborhood of the point $c$, regardless of how small it is. More specifically, "sufficiently close to $c$" means that there exists some $\varepsilon>0$ such that the desired property is true for all $x$ satisfying $0<|x-c|<\varepsilon$.
Note that this condition defines a punctured neighborhood, i.e. not including $c$ itself — this makes sense because we're talking about limits. You can see this phrase a lot in many other different proofs. Depending on the context this same phrase may mean punctured ($0<|x-c|<\varepsilon$) or non-punctured ($|x-c|<\varepsilon$) neighborhood, i.e. without or with $c$ itself.
