You are right in that if $\limsup a_n$ and $\liminf a_n$ are different from each other, then the sequence, as a sequence of real numbers, cannot be convergent, since both $\limsup$ and $\liminf$ are limit points of the sequence, and, by the triangle inequality, a sequence cannot have more than one limit (the uniqueness of the limit of a sequence holds, more generally,for metric spaces,and, even more generally, to Hausdorff spaces).
To throw some more in, $\limsup$ and $\liminf$ are guaranteed to exist in the extended real numbers, since they are the limits of monotone sequences given by the collection of sups and infs; taking sups over increasingly-larger sets gives you a non-decreasing sequence, and similar for the sequence of infs; if these sequences are bounded, then they will converge to a real number, and, if the sequences associated to each are unbounded, then they will converge to $\pm\infty$ respectively.
You can also see $\limsup$ and $\liminf$ as the largest and smallest limit points of the sequence. Finally, to get to your case, I think Wikipedia has a nice rule used to determine what your $\limsup$ and $\liminf$ are: $\limsup a_n$ is the largest value $b$ (to completely ripoff Wikipedia) of $a_n$ such that for each $\epsilon\gt 0$, only finitely many terms of the sequence are larger than $b+\epsilon$. Check to see if: 1) $1$ is a limit point of $\{a_n\}$, i.e., does every neighborhood of $1$ contain infinitely many points of the sequence; and 2) Is it the case that for any $\epsilon>0$ there are only finitely many terms larger than $1+\epsilon$. Do something similar for $a=-1$; check that it is a limit point of $\{a_n\}$, and check also that for every $\epsilon>0$, there are only finitely-many points smaller than $a-\epsilon$.
So, for $1$, for any $\epsilon>0$, there is , by the Archimedean Principle, an integer $N$ with $\frac {1}{N}\lt\epsilon$.
So all terms $1+\frac {1}{n}$ with $n\gt N$ will be smaller than $1+\epsilon$. Then every neighborhood of $1$ contains infinitely many points of $\{a_n\}$, and, by the above paragraph, $1$ is the $\limsup$ of $\{a_n\}$.
Similarly , for $-1$, using the Archimedean principle again, there will be an N' such that $\frac {1}{n} \lt\epsilon$ for all n>N' and so all terms $-1+\frac {1}{n}$, with n>N' will be larger than $-1+\epsilon$.
But notice that your choice of $0$ will not do. Take a neighborhood of $0$ with $\epsilon=\frac {1}{2} $. Then, since both sequences are monotone (because n>n' impies $\frac {1}{n} <\frac {1}{n}$ ) and $a_2= \frac{3}{2}$, while $a_3=\frac{-2}{3}$, there will be no terms at all of the sequence in the neighborhood $(\frac{-1}{2},\frac{1}{2})$ of $0$. So $0$ is not a limit point of $a_n$, let alone $\liminf a_n$.
So, when it comes to $0$: no Sup for you!