Lim (1 + 1/x) ^ (3x) as x->infinity

solving this using L'hopital's rule for indeterminate infinity - infinity.

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