Prove that $\liminf x_n \le \liminf a_n \le \limsup a_n \le \limsup x_n$ Assume that $\{x_n\}$ is a sequence of real numbers and $a_n=\frac{x_1+\dots+x_n}{n}$ .  
a) Prove that $\displaystyle \liminf_{n \to\infty} x_n \le \liminf_{n \to\infty} a_n \le \limsup_{n \to\infty} a_n \le \limsup_{n \rightarrow \infty} x_n$.  
b) Give an example such that all of the limits written above are finite and $\displaystyle \liminf_{n \to\infty} x_n < \liminf_{n \to\infty} a_n < \limsup_{n \to\infty} a_n < \limsup_{n \rightarrow \infty} x_n$.
c) Give an example such that some of the limits written above are finite and some of them are not.  
Note 1 : For a sequence like $\{b_n\}$ we have $\displaystyle \liminf_{n \to\infty} b_n = \lim_{n\to\infty}(\inf\{b_k:k \ge n\})$ and $\displaystyle \limsup_{n \rightarrow \infty} b_n=\lim_{n\to\infty}(\sup\{b_k:k \ge n\})$  
Note 2 : This question is adopted from the book "Real analysis : A first course" written by "Russel Gordon".
Note 3 : A small part of this question is available on this link but my question has a lot more than that.

Partial answer for (a):
First assume that $(x_n)$ is a bounded squence.
Let $L=\limsup_{n\to\infty}x_n<\infty$. By definition of $\limsup$, there exists $K$ such that $x_n<L+\epsilon$ for all $n>K$. (This is the well-known "eventual upperbound" property of limsup.)
Then
$$
\Large\frac{x_1+\dots+x_n}{n}<\frac{x_1+\dots+x_K+(L+\epsilon)(n-K)}{n}
$$
Taking limsup on both sides gives
$\limsup a_n\leq L+\epsilon$
Since $\epsilon>0$ is arbitrary, $\limsup_{n\to\infty}a_n\leq\limsup_{n\to\infty} x_n$.
The case of $\liminf$ should be similar.
$\liminf a_n\leq\limsup a_n$ is automatic (always holds) so you get it for free.
The $(x_n)$ unbounded case, both $\limsup a_n$ and $\limsup x_n$ will be infinite.
