Product of Limitsuperior of bounded sequences $\{a_{n}\}$,$\{b_{n}\}$ are two bounded sequences.How can we prove that $\limsup(a_{n}b_{n})$ = $\lim(a_{n})\limsup(b_{n})$. Is it equal to $\lim(a_{n})\lim(b_{n})$?
If the sequences are not convergent but bounded (for example $a_{n} =(-1)^{n}$ and $b_{n}=(-1)^{n-1}$) How can this result hold.
I have seen the proof for $\limsup(a_{n}b_{n}) \leq \limsup(a_{n}) \limsup(b_{n})$.
Can we say that $\limsup(a_{n})\leq \lim(a_{n})$ ?

In general $\limsup a_nb_n=\limsup a_n\limsup b_n$ does not hold. Just consider the sequences $a_n=(1,0,1,0,\dots)$ and $b_n=(0,1,0,1,\dots)$. In this case $\limsup a_n=\limsup b_n=1$ and $\limsup a_nb_n=0$.
However, if you know that a sequence $(x_n)$ is convergent, then $\limsup x_n=\lim x_n$. So if both $(a_n)$ and $(b_n$) are convergent, then
$$\limsup a_n\limsup b_n = \lim a_n \lim b_n = \lim (a_nb_n) = \limsup (a_nb_n).$$
If you only know that $b_n$ is convergent and $a_n,b_n\ge 0$, then you have 
$$\limsup (a_nb_n)=\limsup a_n\limsup b_n = \limsup a_n\lim b_n.$$
There are already several posts concerning this equality: 


*

*Proof that $\limsup (a_nb_n)=\limsup a_n \lim b_n$

*Checking of a solution to How to show that $\lim \sup a_nb_n=ab$

*Limes superior of product of two sequences
It is also mentioned in the post you linked to: lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $
You can also obtain this from a more general result that
$$\limsup a_n \liminf b_n \le \limsup (a_nb_n) \le \limsup a_n \limsup b_n$$
which holds if you assume that both $a_n,b_n\ge 0$. (Together with using that $\lim b_n=\liminf b_n=\limsup b_n$, if the sequence $(b_n)$ is convergent.)
