... +1},g(M + ε) <yls − j<g(M − ε), if j ∈{k +1, ,2k +1}.(2.17)From (1.2)and(2.17), we haveyls−k = fyls−k−1,xls −2 k−1<fg(M − ε),M − ε = g(M − ε). (2.18)Also (1.2), ... Consider equation xn+1 = 1+xnyn−k,yn+1 = 1+ynxn−k,n = 0,1, , (3.4)where k∈{1,2, } and the initial conditions x−k,x−k+1, ,x0, y−k, y−k+1, , y0∈(0,+∞).Let E = (0,+∞)andf ... (by taking a subsequence) that there existA,D ∈ [m,M]andB,C ∈ [p,P]suchthatlimn→∞xln −1 = A,limn→∞yln−k−1 = B,limn→∞ysn −1 = C,limn→∞xsn−k−1 = D.(2.5)Thus, from (1.2),...