How do you determine the limit of #(3/x^3 - 6/x^6)# as x approaches 0-? Calculus Limits Determining Limits Algebraically 1 Answer Jim H Dec 22, 2016 #3/x^3-6/x^6 = (3x^3-6)/x^6# Explanation: So the limit is #-oo# (Doe not exist because as #xrarr0^-#, #(3x^3-6)rarr-6# and #x^6rarr0^+# 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 1371 views around the world You can reuse this answer Creative Commons License