Global Behavior of the Max-Type Difference Equation xn+1=max{1/xn,An/xn-1}

and Applied Analysis 3 Lemma 2.5. Let {xn}n −1 be a positive solution of 1.1 and limn→∞Pn S. Then S lim supn→∞xn. Proof. Since Pn is a subsequence of xn, it follows that S ≤ lim sup n→∞ xn. 2.6 On the other hand, by xn 1 ≤ Pn for all n ≥ 1, we obtain lim sup n→∞ xn ≤ lim sup n→∞ Pn S. 2.7 The proof is complete. Remark 2.6. Let {xn}n −1 be a positive solution of 1.1 . By Lemma 2.2, we see that if S lim supn→∞xn and xN < S for someN > 0, then xN−1, xN 1 ∈ S, ∞ . For example, if it were xN−1 < S, then it would be PN < S, which would imply lim supn→∞xn < S. Lemma 2.7. Suppose that {xn}n −1 is a positive solution of 1.1 and S lim supn→∞xn. Write ω xn { x : there exist − 1 ≤ k1 < k2 < · · · < kn < · · · such that lim n→∞ xkn x } . 2.8 Then ω xn {S, 1/S}. Proof. If ω xn contains only one point, we may assume by taking a subsequence that Ank → μ < 1 . By taking the limit in the following relationship: xnk 1 max { 1 xnk , Ank xnk−1 } , 2.9