How do you find #lim_(xtooo)log(4+5x) - log(x-1)#? Calculus Limits Determining Limits Algebraically 1 Answer 1s2s2p Apr 11, 2018 #lim_(xtooo)log(4+5x) - log(x-1)=log(5)# Explanation: #lim_(xtooo)log(4+5x) - log(x-1)=lim_(xtooo)log((4+5x)/(x-1))# Using chain rule: #lim_(xtooo)log((4+5x)/(x-1))=lim_(utoa)log(lim_(xtooo)(4+5x)/(x-1))# #lim_(xtooo)(ax+b)/(cx+d)=a/c# #lim_(xtooo)(5x+4)/(x-1)=5# #lim_(uto5)log(u)=log5# 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 1669 views around the world You can reuse this answer Creative Commons License