Finding $\liminf$ and $\limsup$ of $A_n = (\sin (n) -1,\sin (n) + 1)$? For $n \in \mathbb N^*$, $A_n = (\sin (n) -1,\sin (n) + 1)$.
How to show that $\limsup A_n = [-2,2], \liminf A_n = \{0\}$?
What if $A_n = (\sin (n -1),\sin (n + 1))$ instead?
 A: Given a sequence of sets $A_1, A_2, \ldots$, the lim sup of the $A_n$ is the set of points that are seen in infinitely many of the $A_n$, while $\liminf A_n$ is the set of points that are seen in all but a finite number of the $A_n$.
Your sets $A_n$ are intervals of radius $1$ centered on $\sin(n)$. As $n$ varies, the centers of these intervals weave back and forth between $-1$ and $1$ (visiting points in between). I agree that every point between $-2$ and $2$ (exclusive) must be visited infinitely often by your $A_n$, since $\{\sin(n):n=1,2,\ldots\}$ is dense in the interval $[-1,1]$, which implies that for every small $\epsilon>0$ there exists infinitely many $n$ for which $\sin(n)>1-\epsilon$; for those $n$ we have $2-\epsilon<\sin(n)+1$, which means $2-\epsilon\in A_n$. Similarly there are infinitely many $n$ for which $-2+\epsilon\in A_n$. But $\sin(n)$ is never $-1$ or $1$ when $n$ is an integer (since $\pi$ is irrational), so $-2$ and $2$ do not belong to any of the $A_n$. Conclude $\limsup A_n=(-2,2)$.
As for the lim inf, only the point $0$ is found in every $A_n$, since $-1<\sin(n)<1$ for every integer $n$. Any nonzero number won't belong to $\liminf A_n$, because if $\epsilon >0$ then (by the same claim as cited earlier) there are infinitely many $n$ where $\sin(n)>1-\epsilon$. For those $n$ we have $-\epsilon<\sin(n)-1$, which means $-\epsilon\not\in A_n$. Similiarly if $\epsilon <0$ then there are infinitely many $n$ for which $\epsilon\not\in A_n$.
A: Note you only need to know $\limsup \sin n $ and $\liminf \sin n$ to solve this.
The $\limsup$ part has been answered here, and here, the $\inf$ part should be similar.
