Let $(x_{n})$ and $(y_{n})$ be bounded sequences. Prove that $$ \overline{\lim_{n \to \infty}} \biggr( \max \{ x_{n} ,y_{n} \} \biggr) = \max \{ \overline{\lim_{n \to \infty}} x_{n}, \overline{\lim_{n \to \infty}} y_{n} \}$$
Is the following equation also true? $$ \overline{\lim_{n \to \infty}} \biggr( \min \{ x_{n} ,y_{n} \} \biggr) = \min \{ \overline{\lim_{n \to \infty}} x_{n}, \overline{\lim_{n \to \infty}} y_{n} \}$$
My idea is that for the $\geqslant$ proof I should use upper limits monotony and for $\leqslant$ proof subsequences, but I am not sure. Maybe I am on the completely wrong track. To conclude this, after long hours I was not able to prove it myself, so any help would be appreciated.
One inequality is obvious sine $x_n \leq \max \{x_n,y_n\}$ and $y_n \leq \max \{x_n,y_n\}$. For the orther way let $\epsilon >0$ Then there exists $n_1$ such that $x_n < \lim \sup x_n+\epsilon$ for al $n \geq n_1$. Similarly, there exists $n_2$ such that $y_n < \lim \sup y_n+\epsilon$ for al $n \geq n_1$. For $n \geq \max \{n_1,n_2\}$ we get $\max \{x_n,y_n\} <\max \{\lim \sup x_n, \lim \sup y_n\}+\epsilon$. Hence $\lim \sup \max \{x_n,y_n\} <\max \{\lim \sup x_n, \lim \sup y_n\}+\epsilon$. Since $\epsilon$ is arbitrary the proof is complete.
For the counter-example just take $x_n=1$ for $n$ odd, $x_n=0$ for $n$ even and $y_n=1$ for $n$ even, $y_n=0$ for $n$ odd.