Proof on limit superior and limit inferior of a set I understand the result intuitively but how can I prove this?

For a given integral $n \ge 1$, let $A_n  = \left\{\frac mn \mid m \in \mathbb Z\right\}$. Show that $\varlimsup_{n\to\infty} A_n = \mathbb Q$, the set of rational numbers and $\varliminf_{n\to\infty} A_n = \mathbb Z$, the set of integers.

 A: It's obvious that $A_n$ only contains rational numbers, so both limits are contained in $\Bbb Q$.
Limit superior:
We need to show that for any $N \in \Bbb N$, and any rational number $q$ there is an $n \geq N$ such that $q \in A_n$ (i.e. any rational number appears in infinitely many of the $A_n$). Write $q = \frac{a}{b}$ for some $a, b \in \Bbb Z$, with $b$ positive. Can you see why we have $q \in A_{bN}$? Clearly, $bN \geq N$, so the result is proven.
Limit inferior:
We need to show that for any $N \in \Bbb N$, and any rational, non-integer number $q$, there is an $n \geq N$ such that $q \notin A_n$ (i.e. that any non-integer rational number is "missed" by an infinite number of $A_n$). With the same notation as above, can you see why we have $q \notin A_{bN + 1}$? This proves that $\underline\lim_{n \to \infty} A_n \subseteq \Bbb Z$. It is clear that each of the $A_n$ contain all integers, so $\underline\lim_{n \to \infty} A_n \supseteq \Bbb Z$, and the result is proven.
A: Hint. Write down the definition of $\varlimsup/\varliminf A_n$, i. e. 
$$ \varliminf A_n = \bigcup_{k} \bigcap_{n\ge k} A_n, \qquad\varlimsup A_n = \bigcap_k \bigcup_{n\ge k} A_n.  $$
Show the equality of sets as you almost ever do, by showing both inclusions. 
Let me show you one inclusion as guiandance: We will prove $\mathbb Q \subseteq \varlimsup A_n$. So let $q = \frac{m'}{n'} \in \mathbb Q$. We want to prove $$q \in \varlimsup A_n \iff \forall k \, \exists n \ge k : q \in A_n $$
So let $k$ be given, then as $q = \frac{m'k}{n'k}$, we have $q \in A_{n'k}$, so set $n := n'k \ge k$. Hence $q \in \varlimsup A_n$.
I'm sure you can do the other inclusions for yourself.
