The order of $\sqrt{\epsilon(1-\epsilon)}$ and $4\pi^2\epsilon$ as $\epsilon \rightarrow 0$? I was reading on the big O/little O notation etc. and I understand the definitions, but how exactly would I use it to find the order of an expression/function?
I am asked to determine the order of $\sqrt{\epsilon(1-\epsilon)}$ and $4\pi^2\epsilon$ as $\epsilon \rightarrow 0$. But how would I do that?
 A: For $\sqrt{\epsilon(1-\epsilon)}$,
note that $(1-\epsilon)$ isn't small at all
as $\epsilon \to 0$,
so all that matters is the $\epsilon$.
For $4\pi^2\epsilon$,
the $4 \pi^2$
is just a constant,
so it does not affect the
order of the zeroness.
A: Big-Oh means that, simply speaking, asymptotically, $ \forall \epsilon <  \epsilon' \ \exists \ C>0$ s.t. the ratio of the functions $f,g$ are upper-bounded by $C$. In such case we write $f=O(g)$. In your case:
$$
\lim_{\epsilon \to 0}\frac{\sqrt{\epsilon(1-\epsilon)}}{4 \pi^2 \epsilon} = \frac{1}{4 \pi^2} \lim_{\epsilon \to 0}\sqrt{\frac{\epsilon(1-\epsilon)}{\epsilon^2}} = \infty
$$
this means that there are no such $ \epsilon', \ C$ that $ \forall \epsilon <\epsilon' \ f(\epsilon) \leq C g(\epsilon)$. Since the limit of the ratio is $\infty$, we write $f=\omega(g)$ or equivalently $g=o(f)$.  
A: You may do as in the following way:
$$\frac{(\epsilon(1-\epsilon))^{1/2}}{\epsilon^p}=\epsilon^{{0.5}-p}(1-\epsilon)^{0.5}$$
Which goes to zero when $\epsilon$ goes to zero iff $p<0.5$  in this case is $o(\epsilon)$. When $p=0.5$ we have $O(\epsilon)$. The other case is analogous.
