How do you evaluate the limit #(root3x-1)/(sqrtx-1)# as x approaches #1#? Calculus Limits Determining Limits Algebraically 1 Answer Cesareo R. Aug 4, 2016 #2/3# Explanation: #(root3x-1)/(sqrtx-1) equiv (y^2-1)/(y^3-1)# by doing the substitution #y = root6 x# but #(y^2-1)/(y^3-1) =( (y+1)(y-1))/(y^3-1) = (y+1)/(y^2+y+1)# so #lim_{x->1}(root3x-1)/(sqrtx-1) equiv lim_{y->1} (y+1)/(y^2+y+1) = 2/3# 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 1559 views around the world You can reuse this answer Creative Commons License