How do you find the limit of #(sin 2x)/(sin 3x)# as x approaches 0? Calculus Limits Determining Limits Algebraically 1 Answer Shwetank Mauria Mar 28, 2016 #L_(x->0)(sin2x)/(sin3x)=2/3# Explanation: We know that #L_(theta->0)(sintheta)/theta=1# Hence #L_(x->0)(sin2x)/(sin3x)# = #L_(x->0)[(sin2x)/(2x)((3x)/sin(3x))xx2/3]# = #(L_(x->0)(sin2x)/(2x))/(L_(x->0)(sin3x)/(3x))xx2/3# = #1/1xx2/3=2/3# 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 2519 views around the world You can reuse this answer Creative Commons License