Given a sequence $x_k = \sin(\frac{kπ}{2}) + \frac{2k}{(2-(-1)^k}$
Given $a_k = x_{2k-1}$ and $b_k = x_{2k}$
Find $\liminf_{k \to \infty} a_k$ ,
$\limsup_{k \to \infty} a_k$ ,
$\liminf_{k \to \infty} b_k$ and
$\limsup_{k \to \infty} b_k$
What I did was list out the first few characters: $a_n = \frac{5}{3}, 1, \frac{13}{3}, \frac{11}{3},...$ and $b_k = 4, 8, 12, 16,...$
So is it safe to say that
$\liminf_{k \to \infty} a_k = 1$ ,
$\limsup_{k \to \infty} a_k = \infty$,
$\liminf_{k \to \infty} b_k = 4$ and
$\limsup_{k \to \infty} b_k = \infty$
No, your calculations are not right. The $\liminf$ is the smallest number to which a subsequence of $a_k$ or $b_k$ converges. Now it is easy to see that $b_k = 4k$ and this sequence is unbounded and has no convergent subsequence. Therefore we have $$\liminf_{k \to \infty} b_k = \limsup_{k \to \infty} b_k = \infty$$ (as we could also check by the definition $\liminf_{k \to \infty} b_k = \lim_{k \to \infty} \inf_{m \ge k} b_k$ because this infimum will always be $4k$ and the limit of $4k$ as $k \to \infty$ is $\infty$).
Now we also obtain that $a_k = x_{2k-1} = (-1)^{k+1} + \frac{2}{3}(2k-1)= -\frac{2}{3} + (-1)^{k+1} + \frac{4}{3}k$.
As we can see the term $- \frac{2}{3}$ does not depend on $k$. We obtain two subsequences of $a_k$, namely $$a_{2k} = -\frac{2}{3} - 1 + \frac{4}{3} \cdot 2k$$ $$a_{2k-1} -\frac{2}{3} + 1 + \frac{4}{3} \cdot (2k-1)$$ Both these sequences tend to $\infty$, so does the whole sequence $a_k$ (as the two subsequences combined form the whole sequence) and we also get $$\liminf_{k \to \infty} a_k = \limsup_{k \to \infty} a_k = \infty$$
The problem is that $\liminf$ is NOT the smallest number that comes up in the sequence but the smallest number $\xi$ for which a subsequence, say $(a_{n_k})_k$ exists with $\lim_{k \to \infty} a_{n_k} = \xi$