Let's show that $\liminf A_n = C_0$ where $C_0 = \{(x,y) : x^2+y^2<1\}$ and $\bar{C}_0 = C_0 \cup \partial C_0$.
Observe that $\liminf A_n$ can be interpreted as the set of the $w \in \mathbb{R}^2$ such that exist a $n$ such that $w \in A_m$ to all $m \ge n$. So, let $w=(w_1,w_2)$ a point of $C_0$. Can you see that the circles "converges" to $C_0$? I mean, can you convince yourself that for this $w$ exist a $n$ such that $w$ is in all circles $A_m$ for $m$ large enough?
But what happens with the points on the border of $C_0$, well, they can belong to all circles at the same time because the centers are changing his position. If you take $w$ in the border of $C_0$ but on the left of the origin, for large $m$ it won't belong to $A_m$ when $m$ is even...
To the $\limsup A_n$ remember that it is the set of the $w$ which belongs to infinitely many $A_m$'s ...
Hope this can help!
The limiting ratio is not hard to derive via the method of moments.
In a random permutation $\sigma$ of $\{1,2,\dots,n\}$ (not necessarily a derangement), let $X_n$ be
- the total number of integers $i$, $1 \le i \le n$, such that $\sigma(i) = i$, plus
- the total number of integers $i$, $1 \le i \le n-1$, such that $\sigma(i+1) = \sigma(i)+1$, plus
- the total number of integers $i$, $1 \le i \le n-1$, such that $\sigma(i) = \sigma(i+1)+1$.
We'll show that for any fixed $k$, we have $\lim_{n \to \infty} \mathbb E[\binom{X_n}{k}] = \frac{3^k}{k!}.$ If a random variable $X$ is Poisson with mean $3$, we also have $\mathbb E[\binom{X}{k}] = \frac{3^k}{k!}$. The Poisson distribution is determined by its moments, so it follows that $X_n$ converges in distribution to $X$, and in particular $\lim_{n \to \infty} \Pr[X_n = 0] = e^{-3}$.
To see this, note that $X_n$ is the sum of $3n-2$ indicator variables corresponding to each of the events (listed above) that $X_n$ counts, so $\binom{X_n}{k}$ counts the number of size-$k$ sets of events that occur. The calculation is difficult to do exactly but more or less straightforward asymptotically. Out of the $\binom{3n-2}{k}$ choices of $k$ events, $(1-o(1))\binom{3n-2}{k}$ never involve the same value $\sigma(i)$ more than once, so the probability that they occur is $(1+o(1))n^{-k}$. These contribute $(1+o(1))\frac{3^k}{k!}$ to the expected value $\mathbb E[\binom{X}{k}]$, which is all that we wanted.
So it remains to reassure ourselves that the contribution from all other choices of $k$ events is insignificant in the limit. Muddying the picture, some groups of $j \le k$ events that overlap have a significantly higher probability than a group of $j$ nonoverlapping events. For example, the events that $\sigma(3)=3$, that $\sigma(4)=4$, and that $\sigma(4)=\sigma(3)+1$ have this property.
However, we can never win this way, because there are $O(n)$ ways to pick that a group of $j$ overlapping events (with a constant factor depending on $k$) but a $O(n^{-2})$ chance that all of the events occur (since at least two values of $\sigma$ are involved). So for any possible overlapping structure, the contribution to $\mathbb E[\binom{X}{k}]$ is $O(n^{-1})$, and there is a constant number (depending only on $k$) of overlapping structures.
As a result, we can ignore all overlaps, conclude that $\lim_{n\to\infty}\mathbb E[\binom{X}{k}] = \frac{3^k}{k!}$, and deduce that $\lim_{n\to\infty} \Pr[X_n = 0] = e^{-3}$. Therefore $$\lim_{n\to\infty} \Pr[X_n = 0 \mid \text{$\sigma$ is a derangement}] = \lim_{n\to\infty} \frac{\Pr[X_n=0]}{\Pr[\text{$\sigma$ is a derangement}]} = \frac{e^{-3}}{e^{-1}} = e^{-2}.$$
Best Answer
The difference between $\liminf A_n$ and $\limsup A_n$ is very simple : $\limsup A_n$ is all those points that belong to infinitely many $A_n$. But $\liminf A_n$ is all those points that belong to all but finitely many $A_n$.
Another way of putting the $\liminf$ : $x$ doesn't belong in $\liminf A_n$ if(and only if) it doesn't belong in infinitely many of the $A_n$.
To do this, of course we must understand the nature of $A_n$. But that is not difficult : Let $r_n = (-1/n)^n$. We see that $r_n < 0$ for $n$ odd, and $r_n > 0$ for $n$ even, but $r_n \to 0$ as $n \to \infty$. From this , $A_n$ are (open) circles with radius $1$ and center that is converging to $(0,0)$, but the center alternates between the right and the left of the $x$-axis. This information is very useful.
We first run up a triangle inequality corollary : we have $\|x - (r_n,0)\| \geq \|x\| - |r_n|$, very trivially. Using this we can proceed to the trivial cases.
Consider first any $x$ with $\|x\| > 1$. We will show that $x \notin A_n$ for all large enough $n$, but this is clear : if $|r_n| < \|x\|-1$ then $x \notin A_n$ (exercise). Since this holds for large enough $n$ we are done! Consequently, no such $x$ can be in $\limsup A_n$, or therefore in $\liminf A_n$.
Now, we will show that if $\|x\| =1$ then $x \in A_n$ for infinitely many $n$,but $x \notin A_n$ for infinitely many $A_n$ (barring two very special points). Indeed, WLOG let $x$ has positive first coordinate. Then, it is easy to see that if $r_n > 0$ and $|r_n|$ is smaller than the first coordinate of $x$, we have $\|x-(r_n,0)\| < \|x\| = 1$, and if $r_n < 0$ we have $\|x - (r_n,0)\| > \|x\| = 1$ (To see this, write down $x = (x_1,x_2)$ with $x_1>0$ and see what happens). Therefore, $x$ doesn't belong in $\liminf A_n$, but does in $\limsup A_n$.
Of course, if $x$ has negative first coordinate things are similar. But look at $x = (0,1)$ : it doesn't belong in any of the $A_n$! Similarly $(0,-1)$. These two special points don't belong in either the $\limsup$ or $\liminf$ of the sets.
Now, let $\|x\| < 1$. Then there exists $\delta > 1$ such that for all $\|y\|< \delta$ we have $\|x -y\| < 1$. Taking $|r_n| < \delta$, for large enough $n$ it is seen that $\|x - (r_n,0)\| < 1$ so $x \in A_n$. Since this happens for all $n$ after some large enough $N$, we conclude that $x \in \liminf A_n$ and so in $\limsup A_n$.
The characterizations are :
$\limsup A_n = \overline{B(0,1)} \setminus \{(0,1),(0,-1)\}$.
$\liminf A_n = B(0,1)$.