How do you find the limit of #(x^3 - 27) / (x^2 - 9)# as x approaches 3? Calculus Limits Determining Limits Algebraically 1 Answer Shwetank Mauria Mar 30, 2016 #L_(x->3)(x^3-27)/(x^2-9)=9/2# Explanation: #L_(x->3)(x^3-27)/(x^2-9)# = #L_(x->3)((x-3)(x^2+3x+9))/((x-3)(x+3)# = #L_(x->3)(cancel((x-3))(x^2+3x+9))/(cancel((x-3))(x+3)# = #L_(x->3)(x^2+3x+9)/(x+3)# = #L_(x->3)(3^2+3xx3+9)/(3+3)# = #L_(x->3)(9+9+9)/6=27/6=9/2# 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 37216 views around the world You can reuse this answer Creative Commons License