y_n is a bounded sequence
liminf(-y_n)=-limsup(y_n)
n->infinity
i cant understand how it happens??
If $\displaystyle A$ be a non-empty bounded subset of $\displaystyle \mathbb{R}$ then $\displaystyle \sup (-A) = -\inf A \text{ and }\inf (-A) = -\sup A$.
This is the key fact that we need. (Notation: $\displaystyle -A = \{ - a|a\in A\}$)
Let $\displaystyle r = \liminf (-y_n)$.
Therefore, $\displaystyle r = \sup \left\{ \inf\{ -y_k |k\geq n\} | k\geq 1\right\}$
But then, $\displaystyle -r = \inf \left\{ \boxed{- \inf \{ -y_k |k\geq n\}} | k\geq 1\right\}$
Apply again to the boxed expression, $\displaystyle -r = \inf \left\{ \sup \{ y_k |k\geq n\} | k\geq 1 \right\}$
Thus, $\displaystyle -r = \limsup (y_n)$.
Putting this together we get, $\displaystyle \liminf (-y_n) = - \limsup (y_n)$.
You may want to wait for confirmation on this...it may not be rigorous enough....also how late it is may cause concern for incorrectness.
I think you alternatively to TPH's response may be able to say that since $\displaystyle x_n$ is a bounded sequence that the set of subsequential limits $\displaystyle S$ of $\displaystyle x_n$ is a bounded subset of the reals. This fact lets us apply all the basic "rules and theorems" of subsets of the reals to $\displaystyle S$. So now using TPH's first remark and noting that $\displaystyle \limsup x_n=\sup S$ and $\displaystyle \liminf x_n=\inf S$ we can draw our conclusion.