If you replace $n$ with $n^{2}$, this is a sequence of circles that at each stage is increased in radius by $1$ and then shifted to the right by $1$. I claim that $\displaystyle\lim_{n \to \infty} A_{n}$ is actually the right-half plane with the origin, i.e.
$ \{(x,y) \in \mathbb{R}^{2}: x > 0 \} \cup \{0,0\}$
It is easy to see that
$ \{(x,y) \in \mathbb{R}^{2}: x <0 \} \cup \{(0,y): y \neq 0\} $
is not in $\displaystyle\lim_{n \to \infty} A_{n}$. Take $(x,y)$ with positive $x$ coordinate. We wish to find an $N$ large enough so that $(x,y) \in A_{n}$ for $n \geq N$. Without loss of generality, we assume $y$ is positive.
Since the ball is given by
$(x-n)^2 + y^{2} \leq n^2$
we can solve for $y$, to get
$y \leq \sqrt{n^{2} - (x-n)^{2}}$
since $x > 0$, $n^{2} - (x-n)^{2} > 0$
Moreover, its derivative is $2(n-x)$, so that if $n > x$,
$\sqrt{n^{2} - (x-n)^{2}}$
is increasing. Thus, taking $N$ larger than $x$ such that
$y \leq \sqrt{N^{2} - (x-N)^{2}}$
holds, we see that $(x,y) \in A_{n}$ for $n \geq N$.