Combining these two results into one asymptotic notation
Not that I encourage you to do so, but here's one way: 
$$\Theta \left(N\log\left\lfloor\frac{1}{N\epsilon}+1\right\rfloor+ \frac{1}{\epsilon+N^{-1}}+\log N\right) $$

For $\epsilon = \omega(N^{-1})$, the $\log\left\lfloor\frac{1}{N\epsilon}+1\right\rfloor$ expression is $0$, and the left expression is $$\frac{1}{\epsilon+N^{-1}}+\log N = \Theta(\epsilon^{-1} + \log N)$$

For $\epsilon = \Theta(N^{-1})$, the whole expression translates into $\Theta(N)$.

For $\epsilon = o(N^{-1})$ you get:
$$\Theta \left(N\log\left\lfloor\frac{1}{N\epsilon}+1\right\rfloor+ \frac{1}{\epsilon+N^{-1}}+\log N\right) = \Theta\left(N\log\left\lfloor\frac{1}{N\epsilon}+1\right\rfloor+\Theta(N)\right)=\Theta\left(N\log\frac{2}{N\epsilon}\right)$$