Questions about the Fermi-Dirac distribution at $T=0.$ In my book (Daniel V.Schroeder - An introduction to thermal physics) in page 267 they introduce the Fermi-Dirac distribution is given by
$$\bar{n}_{FD}=\frac{1}{e^{(\epsilon-\mu)/kT}+1}. \tag{1}$$
But a few pages later they claim that the Fermi-Dirac distribution becomes a step function and then proceed to define the Fermi energy as
$$\epsilon_F=\mu(T=0).$$
Questions:
1) How does $\bar{n}_{DF}$ become a stepfunction at $T=0$? It's not even possible to plug it in the distribution since then we have division by zero.
2) The expression for the chemical potential $\mu$ is given by
$$\mu=-kT\ln{\frac{Z_1}{N}},$$
where $N$ is the number of particles and $Z_1$ is the partition function for any single particle. So setting $T=0$ should just give $\epsilon_F=\mu=0$. But this is apparently not the case. Why?
 A: 1) Yes. These curves (for example) were computed with Mathematica with $T$ very small, approaching zero but still finite.
Discussion of the limit:
$$ \lim_{T\rightarrow 0} \frac{1}{e^{\delta/T}+1},$$
where $\delta = (E-\mu)/k_B$.
If $E>\mu$, then $\delta >0$ and $\delta/T \rightarrow +\infty$, hence $e^{\delta/T} \rightarrow \infty$ and $f=0$.
If $E<\mu$, then $\delta <0$ and $\delta/T \rightarrow -\infty$, hence $e^{\delta/T} \rightarrow 0$ and $f=1$.
2) I don't know exactly where your formula comes from, but usually you do an expansion with $E_F \gg k_BT$ like in here. That is because you never consider a particle at energy $\epsilon = E_F$, but rather at $\epsilon = E_F + k_B T$.
Also, $Z$ for non-interacting fermions is:
$$ Z = \sum_{n=0}^1 r^n = 1+r,$$
where $r= \exp \left (-\frac{\epsilon - \mu}{k_B T} \right ) $ which thus also depends on temperature.
A: The Fermi-Dirac distribution is a statistical expression for the fact that only a single Fermion can occupy a specific state. Then, the probability of a state to be occupied is given by the expression. Now let's think of a situation where we have a system with $N$ states of energy, ordered from the lowest to the highest as $0 < \epsilon_1 < \epsilon_2 < \cdots < \epsilon_N$. And we have $n$ identical fermions in the system. If the temperature is very low $T \ll \epsilon_1$, then of course the 'first' fermion will have to be in the lowest level, and once it is occupied, the next fermion will have to go to the next level etc. until we will fill the $n$ lowest energy levels. This will give the Fermi energy -- the highest energy occupied when the temperature is very very very low. You can think of $T=0$ as 'temperature is much smaller than the lowest energy scale in the system'. The Fermi-Dirac distribution then becomes effectively a $\theta(E_F-\epsilon)$ function.
Note that this is in contrast to bosons, that can all pile up in the lowest energy level when the temperature is lowered, and therefore there is no concept of Fermi energy for them (on the other hand, you get a cool phenomena like the Bose-Einstein condensate).
Regarding your worry about $\mu = -k T \ln(Z/N)$ leading to a zero chemical potential at zero temperature: this is not the case as $Z$ also depends on temperature. Again - you can expand for small $T$ and get the expression.
