Hi everyone, The function y_1(x)=ln(x) is an unbounded strictly increasing function going to infinity at a slow rate. But, inside it, the function x, goes to infinity faster than ln. So, let's reduce the rate a bit more: y_2(x)=(ln○ln)(x)=ln(ln(x))=ln(y_1(x)) You get it, we can define y_n(x)=(ln○...n-times...○ln)(x) Thus, what is the limit if n goes to infinity? And here's my problem: As it is a compose of unbounded strictly increasing functions, it should be an unbounded strictly increasing function too and should go to infinity as well but, for instance, it never reaches 0. It sounds like a paradox. Case of 0: If x=e(0), y_1(x)=0 If x=e(e(0)), y_2(x)=0 ... If x=(e○...n-times...○e)(0), y_n(x)=0 But lim(x=(e○...n-times...○e)(0),n goes to infinity) is infinity, therefore if n tends to infinity, y_n(x) tends to 0 for this x. And, we can proceed as well for other values of the function. Please help me to correct my approach and to deal with this infinite compose and perhaps to try to answer the question of reaching infinity as slow as possible and by bijection as fast as possible (if there's any answer to this question). Ciao Nicolas