How do you find the limit of #e^x/x^3# as x approaches infinity? Calculus Limits Determining Limits Algebraically 1 Answer Konstantinos Michailidis May 29, 2016 Because it is in the indeterminate form #oo/oo# we can apply L'Hôpital's rule three times respectively to get #lim (e^x/x^3)=lim (((e^x)')/((x^3)'))=lim (e^x)/(3x^2)=(oo/oo)# #lim (((e^x)')/((3x^2)')) = lim (e^x)/(6x)= (oo/oo)# #lim ((e^x)')/((6x)')= lim e^x/6=oo# Finally #lim_(x->oo) e^x/(x^3)=oo# Answer link Related questions How do you find the limit #lim_(x->5)(x^2-6x+5)/(x^2-25)# ? How do you find the limit #lim_(x->3^+)|3-x|/(x^2-2x-3)# ? How do you find the limit #lim_(x->4)(x^3-64)/(x^2-8x+16)# ? How do you find the limit #lim_(x->2)(x^2+x-6)/(x-2)# ? How do you find the limit #lim_(x->-4)(x^2+5x+4)/(x^2+3x-4)# ? How do you find the limit #lim_(t->-3)(t^2-9)/(2t^2+7t+3)# ? How do you find the limit #lim_(h->0)((4+h)^2-16)/h# ? How do you find the limit #lim_(h->0)((2+h)^3-8)/h# ? How do you find the limit #lim_(x->9)(9-x)/(3-sqrt(x))# ? How do you find the limit #lim_(h->0)(sqrt(1+h)-1)/h# ? See all questions in Determining Limits Algebraically Impact of this question 42755 views around the world You can reuse this answer Creative Commons License