Find out the subsequential limits of a sequence 
Let $$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$ and find the upper and lower limits of the sequence $\{x_n\}_{n=1}^\infty$.

We put $n=1, 2, 3 \dots $ then $x_1=-9, x_2 = \frac{15}{2}, \dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.

Notice 
$\underset{n \to \infty}{\lim \sup}\, x_n = \inf \{x_{2n} : n \in \mathbb{N}\} $,
and
$\underset{n \to \infty}{\lim \inf}\, x_n = \sup \{x_{2n-1} : n \in \mathbb{N}\} $.
Moreover, $2$ is a lower bound for the set $\left\{2+\frac{3^{2n}}{(2n)!}+\frac{1}{n^2} : n \in \mathbb{N}\right\}$, and $-2$ is an upper bound for the set $\left\{-2-\frac{3^{2n-1}}{(2n-1)!}-\frac{4}{(2n-1)^2}  : n \in \mathbb{N}\right\} $.
Also observe that we have $\frac{3^n}{n!} < 2^{6-n}$ whenever $n>6$.
Let $\varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
\begin{aligned}&N_1  > 7 - \frac{\log\varepsilon}{\log 2}, \text{ and} \\&
N_2\varepsilon > 8 .
\end{aligned}
Set $N=\max\{7, N_1, N_2\}$. So if $n \geq N$, then we have
\begin{equation}\frac{3^n}{n!}+\frac{4}{n^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .
\end{equation}
So there is $N \in \mathbb{N}$ such that 
\begin{aligned} &x_{2N}<2+\varepsilon, \text{ and} \\& x_{2N-1}>-2-\varepsilon.
\end{aligned}
Therefore, $\underset{n \to \infty}{\lim \sup}\, x_n =2$ and $\underset{n \to \infty}{\lim \inf}\, x_n =-2$.
Note: Since $N \geq 7$, we know that $2N-1>N$ and $2N>N$.
