Suppose that $(a_n)_n$ and $(b_n)_n$ are two real sequences converging to $a$ and $b$ respectively (both in $\mathbb{R}$). Let $a > b$ and define the sequence $(x_n)_n$ as follows: $$x_{2n} = a_n \quad x_{2n-1} = b_n,$$ that is $(x_n)_n = (a_0, b_1, a_1, b_2, \ldots)$.
I am asked to prove that lim inf of $x_n$ is $b$ and limsup equals $a$.
I first noticed that for some $n_0$ we have that $a_n > b_n$ for all $n \geq n_0$. Indeed, the sequence $(a_n - b_n)_n$ converges to $a-b > 0$ and therefore there is an $n_0 \in \mathbb{N}$ such that $$|a_n - b_n - (a-b)| < \frac{a-b}{2}$$ which is equivalent with $$\frac{a-b}{2} < a_n - b_n < 3\frac{a-b}{2}.$$
In class, we defined lim inf as $\lim_{n \to + \infty} y_n$ with $y_n = \inf\{x_k \vert k \geq n\}$. I think it suffices to look at $\inf\{b_k \vert 2k-1 \geq n\}$ (because $a_n > b_n$ for all $n$) to determine the $y_n$ (I still need to explain why I can discard of the $a_n$ in this infimum) and hence it limit, but I am stuck... The same with limsup. I know both exists, since the sequences $(a_n)_n$ and $(b_n)_n$ are both converging, hence bounded.
Can someone give a hint on this problem? Am I working in the right direction?