Logs? ?of logs? Compare the growth rates of ln x, ln (ln x)and ln (ln (ln x))

Solution Step 1 In this problem we have to compare the growth rate of the functions ln x, ln (ln x) and ln (ln (ln x)) Let f(x) = ln x, g(x) = ln (ln x)), h(x) = ln (ln (ln x)). In order to compare the growth rate we will be using the following condition. “If there are two functions f(x)and g(x)both tending to infinity as x . Then the function f grows faster than gas x if the following condition holds f(x) g(x) lim = or lim = 0 ” x g(x) x f(x) l'Hôpital's Rule: f(x) f(x) Suppose that we have one of the following cases, lim g(x)= 0r lim g(x)= ± xa xa Where a can be any real number, infinity or negative infinity. f(x) f(x) In these cases we have lim g(x)= lim gx) xa xa Step 2 First let us consider two functions f(x) = ln xand g(x) = ln (ln x) These two functions tends to as x approaches g(x) Let us find the limit lim f(x) x lim g(x)= lim ln (ln =) x f(x) x ln x So we can apply l’hopital’s rule. ln (ln x) dx(ln (ln x)) xm ln x = x dxln x) d (ln (ln x)) = 1 ( ) dx ln x x d (ln x) = 1 dx x d 1 1 ln (ln x) dxln (ln x)) ln x x Thus lix ln x = x dxln x) = x x = lim 1 = 0 x ln x ln (ln x) Therefore lim ln x = 0 x Hence by using the condition in step 1, we conclude that ln xgrows faster than ln (ln x) ………. ( 1)