What is the limit of # (lnx)^2/x# as x approaches #oo#? Calculus Limits Determining Limits Algebraically 1 Answer Cesareo R. Aug 18, 2016 #0# Explanation: #(log_ex)^2/x = ((log_e x)/sqrt(x))^2# so #lim_{x->oo}(log_ex)^2/x = (lim_{x->oo}(log_e x)/sqrt(x))^2# but #lim_{x->oo}(log_e x)/sqrt(x)=lim_{x->oo}(e^{log_e x})/e^{sqrt(x)} = lim_{x->oo}(x)/e^{sqrt(x)}# Making now #sqrt(x) = y# #lim_{x->oo}(x)/e^{sqrt(x)}equivlim_{y->oo}(y^2)/e^y = 0# 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 2001 views around the world You can reuse this answer Creative Commons License