How do you evaluate the limit #sin^3(2x)/sin^2(3x)# as x approaches #0#? Calculus Limits Determining Limits Algebraically 1 Answer Cesareo R. Aug 26, 2016 #0# Explanation: After the textbook result #lim_(x->0)sin x/x = 1# we have #sin^3(2x)/sin^2(3x) = (sin(2x)/(2x))^3/(sin(3x)/(3x))^2(2x)^3/(3x)^2# then #lim_(x->0)sin^3(2x)/sin^2(3x) = 1/1 8/9lim_(x->0) x = 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 13750 views around the world You can reuse this answer Creative Commons License