4. The sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots$ satisfies $\lim _{n \rightarrow \infty}\left(x_{2 n}+x_{2 n+1}\right)=315$ and $\lim _{n \rightarrow \infty}\left(x_{2 n}+x_{2 n-1}\right)=2003$. Evaluate $\lim _{n \rightarrow \infty}\left(x_{2 n} / x_{2 n+1}\right)$
Solution:
$$
-1
$$

We have $\lim _{n \rightarrow \infty}\left(x_{2 n+1}-x_{2 n-1}\right)=\lim _{n \rightarrow \infty}\left[\left(x_{2 n}+x_{2 n+1}\right)-\left(x_{2 n}+x_{2 n-1}\right)\right]=315-2003=$ -1688 ; it follows that $x_{2 n+1} \rightarrow-\infty$ as $n \rightarrow \infty$. Then
$$
\lim _{n \rightarrow \infty} \frac{x_{2 n}}{x_{2 n+1}}=\lim _{n \rightarrow \infty} \frac{x_{2 n}+x_{2 n+1}}{x_{2 n+1}}-1=-1,
$$
since $x_{2 n}+x_{2 n+1} \rightarrow 315$ while $x_{2 n+1} \rightarrow-\infty$.