What does it mean that a function dominates over another function? Let $f,g$ be functions with Domain $\mathbb{R}$ and Range $\mathbb{C}$ What does it mean that $f$ dominates over $g$? Unfortunately I could not find any ressources online about this Topic. 
But I remember there was a distinction between the behaviour at infinity and at $0$.
I understand why we call $f$ to be dominating over $g$ if 
$\lim_{x\rightarrow x_0}f(x)=\infty$ and $\lim_{x\rightarrow x_0}g(x)=\infty$ and also $\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=\infty$
Because the values for $f$ are bigger than those of $g$
However I don't understand why we also call $f$ to be dominating over $g$ if 
If $\lim_{x\rightarrow x_0} f(x)=0$ and $\lim_{x\rightarrow x_0}g(x)=0$ and also $\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=0$
Because the values of $f$ are smaller than those of $g$.
If we let $f(x)=x^2$ and $g(x)= x$ 
Hope someone can tell me why in both cases we would call $f$ to be dominating over $g$. 
 A: You've right that the second one is backwards. Whoever you got that from mixed it up.
The idea is that if we have a sum $$f(x)+g(x)$$ and factor out the “dominant” term $f(x)$, so that we get
$$
f(x) \left( 1 + \frac{g(x)}{f(x)} \right)
,
$$
then we want $g(x)/f(x)$ to tend to zero, so that the whole parenthesis just tends to $1+0=1$. If not, it wouldn't be right to call $f$ the “dominant” term.
Saying that $f$ dominates over $g$  (as $x$ tends to whatever it tends to) means that the important contribution to the sum is the one that we get from $f(x)$, whose absolute value is large in comparison to the absolute value of $g(x)$.
So regardless of whether $x \to \infty$ or $x \to x_0$, we say that $f(x)$ dominates over $g(x)$ if
$$
\frac{g(x)}{f(x)} \to 0
.
$$
(The values of $f(x)$ and $g(x)$ may both be “small”, or both “large”, or maybe one small and one large, but the quotient $g(x)/f(x)$ should be small.)
The distinction between $x \to \infty$ and $x \to 0$ that you seem to remember might be that for $x \to \infty$, higher powers $x^k$ dominate over lower powers, but as $x \to 0$ it's the other way around.
A: Colloquially, for "dominating", think "more powerful".
In the case of $\lim_{x\to x_0} f(x)/g(x) = \infty$, think of it as "$f$ goes to infinity more powerfully than dividing by $g$ can pull it toward $0$."
In the case of $\lim_{x\to x_0} f(x)/g(x) = 0$, think of it as "$f$ goes to $0$ more powerfully than dividing by $g$ blows it up to infinity."
A: $\lim_{x \rightarrow x_o} \frac{f(x)}{g(x)} = \lim_{x\rightarrow x_o} f(x) \cdot \lim_{x\rightarrow x_o} \frac{1}{g(x)}$
The first limit goes to $0$ and the second goes to $\infty$, for the overall limit be $0$ the first has to go to $0$ faster than the second limit goes to $\infty$, so the $f(x)$ is the one who drives the limit,, the $f(x)$ dominates $g(x)$.
