What is # lim_(x->-oo) x^2/(x^2-7) #? Calculus Limits Determining Limits Algebraically 1 Answer VNVDVI May 11, 2018 #1# Explanation: We can divide both numerator and denominator by #x^2:# #lim_(x->-oo)((cancel(x^2))/(cancel(x^2)))/(cancel(x^2)/cancel(x^2)-7/x^2)# #=lim_(x->-oo)1/(1-7/x^2)=1/(1-7/(-oo)^2)=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 2360 views around the world You can reuse this answer Creative Commons License