That Is Close 3

From the given limits, we also find $\lim_{n\to \infty}(x_{2n+2}-x_{2n})=1688\\ \lim_{n\to \infty}(x_{2n+1}-x_{2n-1})=-1688$ Thus, for any $\epsilon>0$ , for sufficiently large $n$ , we have $x_{2n+2}=x_{2n}+1688+a_n \\ x_{2n+3}=x_{2n+1}-1688+b_n$ where the real sequences $\{a_n\},\{b_n\}$ are such that, for sufficiently large $n$ , $|a_n|,|b_n|<\epsilon$ . Roughly speaking, the sequences $\{x_{2n}\}$ and $\{x_{2n+1}\}$ are more or less "asymptotically" arithmetic progressions, with common differences $1688,-1688$ respectively. Thus, starting from a sufficiently large number $N$ , we see that $\forall n\ge N$ , $\frac{x_{2n+1}}{x_{2n+3}}=\frac{x_{2N}+1688(n-N)+\sum_{k=N}^n a_k}{x_{2N+1}-1688(n-N)+\sum_{k=N}^n b_k}$ So that, as $n\to \infty$ $\lim\sup_{n}\frac{x_{2n+1}}{x_{2n+3}}\le \lim_{n\to \infty}\frac{x_{2N}/(n-N)+1688+\epsilon}{x_{2N+1}/(n-N)-1688-\epsilon}=\frac{1688+\epsilon}{-1688-\epsilon}=-1$ and $\lim\inf_{n}\frac{x_{2n+1}}{x_{2n+3}}\ge \lim_{n\to \infty}\frac{x_{2N}/(n-N)+1688-\epsilon}{x_{2N+1}/(n-N)-1688+\epsilon}=\frac{1688-\epsilon}{-1688+\epsilon}=-1$ Thus the limit is $\boxed{-1}$ .