Determine $a_n :=\frac{2^n+(-3)^n}{(-2)^n+3^n}, \lim \inf a_n,\lim \sup a_n, \lim a_n $. I am to determin $a_n :=\frac{2^n+(-3)^n}{(-2)^n+3^n}, \lim \inf a_n,\lim \sup a_n, \lim a_n.$
I was checking the sequence for the very first elements:
$n=1$: $\;\frac{2^1+(-3)^1}{(-2)^1+3^1}=-1:$
$n=2: \;\frac{2^2+(-3)^2}{(-2)^2+3^2}=1.$
$n=3$: $\;\frac{2^3+(-3)^3}{(-2)^3+3^3}=-1.$
$n=4$: $\;\frac{2^4+(-3)^4}{(-2)^4+3^4}=1.$
$n=5$: $\;\frac{2^5+(-3)^5}{(-2)^5+3^5}=-1.$
So I assume that for odd $n$ values $a_n = -1$ and for even $n$ values $a_n = 1.$
My assumption is that $\lim \inf a_n = -1$,$\lim \sup a_n = 1$ and the $\lim a_n$ doesn't exist, but I'm struggling at this point. I would appreciate your help a lot.

Note that if $n$ is even, then $(-2)^n = (-1)^n 2^n = 2^n$; similarly, $(-3)^n = (-1)^n 3^n = 3^n$.  Hence $$a_n = \frac{2^n + (-3)^n}{(-2)^n + 3^n} = 1, \quad n \text{ even}.$$
And if $n$ is odd, then $(-2)^n = -2^n$, and $(-3)^n = -3^n$, hence $$a_n = \frac{2^n - 3^n}{-2^n + 3^n} = -1, \quad n \text{ odd}.$$
Consequently, for all integers $n$,
$$a_n = \begin{cases}1, & n \text{ even} \\ -1, & n \text{ odd.} \end{cases}$$  We can also write this simply as $a_n = (-1)^n$.
Thus, $$\sup a_n = 1, \quad \inf a_n = -1,$$ and $$\liminf a_n = 1, \quad \liminf a_n = -1.$$  Since $\limsup a_n \ne \liminf a_n$, it follows that $\lim a_n$ does not exist.
How do we justify the supremum and infimum?  Well, we established $a_n \le 1$ for all $n$, and $a_2 = 1$ implies that this bound is tight.  A similar reasoning applies for the infimum.  And since $a_{2m} = 1$ for all positive integers $m$, the limit superior is $1$, and similar reasoning applies for the limit inferior.
