Does this number blow up or go to zero? $2\pi \epsilon^{3}(\log(\epsilon)+2\pi)^2$
Let $\epsilon \to 0$.  I believe $2\pi \epsilon^{3}(\log(\epsilon)+2\pi)^2 \to 0$, and there is no concern about the blow-up of log at the origin.  But can I be more specific and say that this number goes to zero like $\epsilon^3$?  I.e., this number is in O($\epsilon^3)$.
What if the exponent on log is very large (but fixed)?  Does it matter?  
Thanks,

Given any fixed $t > 0$, for $\epsilon$ small enough you will have $|\ln(\epsilon)| < \epsilon^{-t}$, which ensures your limit is zero and that it goes to zero faster that $\epsilon^{3 - \delta}$ for any positive $\delta$ (but not $\delta = 0$).
One way to see the above inequality  is to write $\epsilon = 1 / x$ and apply l'Hopital's rule on ${\displaystyle {\ln x \over x^t}}$ as $x \rightarrow \infty$.
