How do you find the limit of #x/(sqrt(4x^2 - x + 3))# as x approaches negative infinity? Calculus Limits Determining Limits Algebraically 1 Answer Konstantinos Michailidis Mar 12, 2016 It is #lim_(x->-oo) x/(sqrt(4x^2 - x + 3))= lim_(x->-oo) x/[absx*sqrt[4-1/x+3/x^2]]= lim_(x->-oo) x/[(-x)*sqrt[4-1/x+3/x^2]]= lim_(x->-oo) -1/[sqrt[4-1/x+3/x^2]]= lim_(x->-oo) -1/[sqrt(4-0+0)]= =-1/2# Note that as #x->-oo# , #absx=-x# and #1/x->0# , #1/x^2->0# 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 3529 views around the world You can reuse this answer Creative Commons License