lim ln([(x+1)/x]^3x) as x ->.infinity =lim ln([(x+1)^(3x)]/[x^(3x)]) as s->infinity =lim ln((x+1)^(3x))-ln(x^(3x)) = infinity - infinity
your answer is e3 but you can use l'hopital if you liketake the log, get 3xln(1+1/x)which is in the form ∞×0 then use the usual trick of rewriting as ln(1+1/x)/1/3x
