Physical Intuition for thermodynamic limit of the average occupancy, $\langle N\rangle$, of an orbital 
Say your allowed occupancies for an orbital are $0, 1, 2$, and the values of energy associated with these occupancies are $0, \epsilon, 2 \epsilon$.
Give an expression for the ensemble average occupancy $\langle N\rangle$ when the system composed of this orbital is in thermal and diffusive contact with a reservoir at temperature $\tau$ and chemical potential $\mu$.
How does $\langle N\rangle$ behave in the limit:
(a) $\tau \to \infty$
(b) $\tau \to 0$ with $\mu < \epsilon$
(c) $\tau \to 0$ with $\mu > \epsilon$

I found $\langle N\rangle$ to be:
$$\langle N\rangle = \frac{\exp\left(\frac{\mu-\epsilon}{\tau}\right) + 2 \exp\left(\frac{2(\mu-\epsilon)}{\tau}\right)}{1 + \exp\left(\frac{\mu-\epsilon}{\tau}\right) + \exp\left(\frac{2(\mu-\epsilon)}{\tau}\right)}$$
And am reasonably confident it is correct. This means that
as $\tau \to \infty$, $\space \langle N\rangle \to 1$
as $\tau \to 0$, $\space \langle N\rangle \to 0 \space$ for $\mu < \epsilon$
as $\tau \to 0$, $\space \langle N\rangle \to 2 \space$ for $\mu > \epsilon$
But why should I expect these answers to be correct?
I believe that as $\tau \to 0$, the particles will enter the lowest energy state available to them, and as far as I understand chemical potential, it is sort of a measure for how much work must be done to remove a particle from a system.
However, these do not seem align with the answers I got, so perhaps I am thinking about this wrong.
 A: 
I believe that as τ→0, the particles will enter the lowest energy state available to them, and as far as I understand chemical potential, it is sort of a measure for how much work must be done to remove a particle from a system.

You should think of the system as being in contact with a particle reservoir. If I get rid of a particle (let it flow out to the reservoir) it is sort of like adding an amount energy $\mu$.
If $\mu$ if less than $\epsilon$ the particle would rather be in the reservoir than the orbital.
At zero temperature, if $\mu < \epsilon$, the lower energy state is to just have the particles go out to the reservoir and not populate the orbital (of energy $\epsilon$ at least if populated). So $<N>$ is zero.
If $\mu > \epsilon$, the lower energy state is to fully populate the orbital with two particles (each having energy $\epsilon$ so the total energy is $2\epsilon$ instead of $2\mu$). So $<N>$ is two.
A: There is not only thermal contact in this problem but also chemical. Hence the particles will enter the state with lowest $E - \mu N$ value as $\tau \to 0$.
