limit inf /sup if $x_n\leq y_n$ I have just 2 problems :
1)
Find the $\limsup x_n$ and $\liminf x_n$  where $x_n= e^{-n}$.
2)
Let $x_n\leq y_n$  for every $n\in\mathbb{N}$   .
Show that $\liminf x_n\leq\liminf y_n$ and $\limsup x_n\leq\limsup y_n$ 
thanks..
 A: I will assume that we work with the following definitions:
$$
\limsup x_n = \lim_{n\to\infty} \sup\{x_k; k\ge n\}\\
\liminf x_n = \lim_{n\to\infty} \inf\{x_k; k\ge n\}
$$
(There are several equivalent definitions of limit superior and limit inferior; different authors prefer different definitions.)
We know that the above limits exists (but may be infinite) for any real sequence $(x_n)$.

The sequence $x_n=e^{-n}$ converges to $0$. If a sequence has a limit, then both $\limsup$ and $\liminf$ of this sequence are equal to this limit.
If a sequence $x_n$ converges to limit $L$ then there is an $n_0$ such that $|x_n-L|<\varepsilon$ for $n\ge n_0$, which is the same as saying that
$$L-\varepsilon < x_n < L+\varepsilon \qquad \text{for }n\ge n_0.$$
This implies that for $n\ge n_0$ we have $\sup\{x_k; k\ge n\}\le L+\varepsilon$ and $\inf\{x_k; k\ge n\}\ge L-\varepsilon$; and thus
$$\newcommand{\ve}{\varepsilon}L-\ve\le \liminf x_n \le \limsup x_n \le L+\ve.$$
These inequalities are true for any $\ve>0$, therefore we get $L=\liminf x_n=\limsup x_n$.

Now if we assume that $x_n\le y_n$ for each $n$, then we also have 
$$\sup\{x_k; k\ge n\} \le \sup\{y_k; k\ge n\}$$
for each $n$. Taking limit $n\to\infty$ we get
$$\limsup x_n \le \limsup y_n.$$
