## amc-2014-sr-20

Given that $$f_1(x)=\displaystyle\frac x{x+1} %speech% ,f 1 of x equals x over x plus 1,$$ and $$f_{n+1}(x)=f_1(f_n(x)) %speech% ,f n plus 1 of x equals f one of f n of x,$$, then $$f_{2014}(x) %speech% ,f 2014 of x,$$ equals