I understand that the usual way of deriving the Boltzmann distribution involves considering a small system of energy $\epsilon$ embedded in a much larger heat bath of energy $E - \epsilon$ and the total system energy is $E$.
Given that the bath has accessible microstates $\Omega_b(E - \epsilon)$ and the system has microstates $\Omega_s(\epsilon)$ and the total system + bath has microstates $\Omega_t(E)$.
We now state that the probability of finding the system in energy $\epsilon$ is simply \begin{equation} P(\epsilon) = \frac{\Omega_s(\epsilon)\Omega_b(E - \epsilon)}{\Omega_t(E)} \end{equation}
$\textbf{Question 1}$: This is usually replaced by \begin{equation} P(\epsilon) = \frac{\Omega_b(E - \epsilon)}{\Omega_t(E)} \end{equation}
Why should this be so? Why can we ignore the fact that the number of accessible microstates in the system that should vary with $\epsilon$?
Moving on, the usual trick is to use logarithms, since the $\Omega$ terms are all very large. Thus, \begin{equation} \ln P(\epsilon) = \ln\Omega_b(E - \epsilon) -\ln \Omega_t(E) \end{equation}
A Taylor expansion of the first term about $E$ gives us \begin{equation} \ln P(\epsilon) = c - \epsilon\frac{\partial\ln\Omega_b(E)}{\partial E} + O(\epsilon^2), \end{equation}
where c is just a constant that depends on the bath and will be fixed when we normalize $P(\epsilon)$. Ignorning the higher order terms, we get $P(\epsilon) \propto e^{-\beta\epsilon}$ where we identify $\beta = \frac{\partial\ln\Omega(E)}{\partial E}$.
$\textbf{Question 2}$: Why can we ignore the higher order Taylor terms? I understand that $\epsilon$ is small compared to the heat bath energy $E$ but why am I comparing it to $E$ to get the right measure of the error in probability?