How do you evaluate the limit #(x^4+3^x)/(x^5+1)# as x approaches #oo#? Calculus Limits Determining Limits Algebraically 1 Answer Eddie Aug 29, 2016 #oo# Explanation: #lim_(x to oo) (x^4+3^x)/(x^5+1)# this is in #oo/oo# indeterminate form and therefore L'Hopital's rule applies on repeated basis as follows #= lim_(x to oo) (4x^3+ln (3) 3^x)/(5x^4)# #= lim_(x to oo) (12x^2+ln^2 (3) 3^x)/(20x^3)# #= lim_(x to oo) (24x+ln^3 (3) 3^x)/(60x^2)# #= lim_(x to oo) (24+ln^4 (3) 3^x)/(120x)# #= lim_(x to oo) (ln^5 (3) 3^x)/(120)# #= infty# 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 1429 views around the world You can reuse this answer Creative Commons License