I know that the Gibbs distribution at a particular state, x, is given by $\frac{e^{-\beta E_i}}{\sum_j e^{-\beta E_j}}$ with $\beta = \frac{1}{T}$, but I do not understand what a limiting distribution means or what $\mu(x)$ is. I know that $\mu(x)$ represents the invariant distribution.
By limiting distribution, they mean the distribution to which the Gibbs distribution converges as you take the limit $T\to0$ or $T\to\infty$.
The limit $T\to\infty$ is a bit easier to take, since in this case $\beta\to0$, so you can calculate the limiting distribution directly by substituting $\beta=0$ into the expression you wrote; you don't need any information about the $E_i$ in this case.
The limit $T\to0$ corresponds to $\beta\to\infty$. To find this, divide your expression through by $\mathrm e^{-\beta E_{\text{min}}}$. Then all exponentials have non-positive exponents, so you can take their limit for $\beta\to\infty$.
That leaves the question which state(s) have minimal energy. Note that your expression decreases strictly monotonically with increasing energy. That allows you to identify the state(s) with minimal energy from the given distribution at finite $T$.