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.

Here is a less rigorous but quicker way to look at the problem.
We have 
$$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$
Let's look at 
$$\lvert x_n\rvert=2+\frac{3^n}{n!}+\frac{4}{n^2}$$
We know that
$\dfrac4{n^2}$ is decreasing and $2$ is constant.
The first few terms of $\dfrac{3^n}{n!}$ are
$$\dfrac{3}{1},\,
\dfrac{3\cdot3}{2\cdot1},\,
\dfrac{3\cdot3\cdot3}{3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3}{4\cdot3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3\cdot3}{5\cdot4\cdot3\cdot2\cdot1},\cdots$$
which has 2nd and 3rd terms equal and 4th terms onwards decreasing.
All in all, $\lvert x_n\rvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.
