Is there a sequence $x_n\to+\infty$ such that $\liminf x_{2n}/x_n = 0$?
Hint: If
$$ x_n = \begin{cases}n, & \text{if $n$ is even,} \\ n^2, & \text{if $n$ is uneven,} \end{cases}$$
what's the limit of
$$\frac{x_{2(2n+1)}}{x_{2n+1}}\;\; ?$$
However, it is indeed impossible that $\frac{x_{2n}}{x_n}$ itself converges to zero. The proof given by mechanodroid is entirely correct. For the problem with taking subsequences, see my comment below mechanodroid's answer.
Answer to the original question whether it is possible that $x_n \to +\infty$ but $\frac{x_{2n}}{x_n} \to 0$.
Assume that $\frac{x_{2n}}{x_n} \to 0$. Then there exists $n_0 \in \mathbb{N}$ such that $n \ge n_0 \implies \left|\frac{x_{2n}}{x_n}\right| \le 1$ or $|x_{2n}| \le |x_n|$.
Therefore $$|x_{n_0}| \ge |x_{2n_0}| \ge |x_{4n_0}| \ge \cdots$$
so $(x_{2^kn_0})_k$ is a (by absolute value) decreasing subsequence of $(x_n)_n$ so $x_n \not\to +\infty$.
Tentative answer to whether it is possible that $x_n \to +\infty$ but $\liminf_{n\to\infty}\frac{x_{2n}}{x_n} \to 0$
Assume that $\liminf_{n\to\infty} \frac{x_{2n}}{x_n} \to 0$. Then there exists a subsequence $(x_{p(n)})_n$ such that $\frac{x_{2p(n)}}{x_{p(n)}} \to 0$ so by the previous part there is a further subsequence of $(x_n)_n$ which is decreasing so $x_n \not\to +\infty$.
However, this is wrong because the constructed decreasing subsequence is not necessarily a subsequence of $(x_{p(n)})_n$.