Given #(sin x)^(3x^2)# how do you find the limit as x approaches 0? Calculus Limits Determining Limits Algebraically 1 Answer Cesareo R. Oct 7, 2016 #1# Explanation: #(sin x)^(3x^2) = (1+sinx - 1)^(3x^2) =# #=1+3x^2(sinx-1)+3x^2(3x^2-1)(sinx-1)^2/(2!)+cdots=# #1+x^2f(x)# so #lim_(x->0)(sin x)^(3x^2) = 1+lim_(x->0)x^2 cdot lim_(x->0)f(x) = 1 +0 cdot 0 = 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 1294 views around the world You can reuse this answer Creative Commons License