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 3803 views around the world You can reuse this answer Creative Commons License