How do you find the limit of #(ln x^2) / (x^2-1)# as x approaches 1? Calculus Limits Determining Limits Algebraically 1 Answer Cesareo R. Jul 1, 2016 #lim_{x->1}(log_e x^2) / (x^2-1)=1# Explanation: #(log_e x^2) / (x^2-1) = log_e x/(x-1)-log_e x/(x+1)# but the #log_e x# series expansion in the #x=1# vicinity is #log_e x = sum_{k=1}^{oo}((x-1)^k)/k# so #log_e x/(x-1) = 1 + sum_{k=2}^{oo}((x-1)^{k-1})/k# so #lim_{x->1}log_e x/(x-1) = 1# Finally #lim_{x->1}(log_e x^2) / (x^2-1) = lim_{x->1}log_e x/(x-1)-lim_{x->1}log_e x/(x+1) = 1 + 0=1 # 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 3608 views around the world You can reuse this answer Creative Commons License