How do you evaluate the limit #(w+6)/(w^2+8w+16)# as w approaches #-3#? Calculus Limits Determining Limits Algebraically 1 Answer Alan P. Oct 6, 2016 #lim_(wrarr-3)(w+6)/(w^2+8w+16) = color(green)(3)# Explanation: Since #w^2+8w+6 != 0# when #w=-3# #lim_(wrarr-3)(w+6)/(w^2+8w+16)# #color(white)("XXX") = ((-3)+6)/((-3)^2+8 * (-3) +16)# #color(white)("XXX") = 3/1 = 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 1427 views around the world You can reuse this answer Creative Commons License