How do you find the limit of #ln((2x)/(x+1))# as x approaches infinity? Calculus Limits Determining Limits Algebraically 1 Answer Eddie Aug 22, 2016 # = ln 2# Explanation: #lim_(x to oo) ln((2x)/(x+1))# as #ln# is continuous we can say: #=ln ( lim_(x to oo) (2x)/(x+1) )# #=ln ( lim_(x to oo) (2)/(1+1/x) )# and as #lim_(x to oo) 1/x = 0# #=ln ((2)/(1+0) ) = ln 2# 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 4727 views around the world You can reuse this answer Creative Commons License