Big-O, Omega, Theta and Orders of common functions 
Based on this table, is it generally going to be true that for two functions whose most "significant" terms are of the same order that they will be big-Theta each other? And a function of order lower on this table will be big-O ones above it?
Here is the context of my question if it helps:

 A: I'm not sure I fully understand your question, but have a look at (b):
$$
\lim_{n \to \infty} \frac{\sqrt{n}}{n^{\frac{2}{3}}} = 0
$$
This means $f(n) = o(g(n))$, i.e. we can pick $c >0$ arbitrarily small, and $\forall c>0 \ \exists \ N(c)\ s.t. \ \forall n>N(c) \ \sqrt{n} < c n^{\frac{2}{3}}$. At the same time $\nexists N(s) \ s.t. \forall n>N(s) \ \sqrt{n} >s n^{\frac{2}{3}}$.Hence $f(n) \neq \Omega (g(n))$. 
Here $f(n)$ is the numerator and $g(n)$ is the denominator.
A: 
is it generally going to be true that for two functions whose most "significant" terms are of the same order that they will be big-Theta each other?

Yes, in the following sense. 
Assume that $f(n)=h(n)+r(n)$ and $g(n)=k(n)+s(n)$ with $h(n)=\Theta(k(n))$, $r(n)=o(h(n))$ and $s(n)=o(h(n))$. Then indeed $f(n)=\Theta(g(n))$.
A: It is not true that all functions corresponding to the same entry in the table will be $\Theta$ of each other.  In particular, $a^x = O(b^x)$ if and only if $|a| \leq |b|$.  I'm not sure about the L-notation bit.
It is true, however, that $f = O(g)$ if $g$ is lower down on the table.
