Question #20e44 Calculus Differentiating Trigonometric Functions Limits Involving Trigonometric Functions 1 Answer Cesareo R. Apr 16, 2017 #oo# Explanation: #(x^2+sin^2x+xsinx)/sin^3x = 1/sinx(x^2/sin^2x+1+x/sinx)# so #lim_(x->0)(x^2+sin^2x+xsinx)/sin^3x=lim_(x->0)(1/sinx)lim_(x->0)(x^2/sin^2x+1+x/sinx)=1/0(3)=oo# Answer link Related questions How do you find the limit of inverse trig functions? How do you find limits involving trigonometric functions and infinity? What is the limit #lim_(x->0)sin(x)/x#? What is the limit #lim_(x->0)(cos(x)-1)/x#? What is the limit of #sin(2x)/x^2# as x approaches 0? Question #99ee1 What is the derivative of #2^sin(pi*x)#? What is the derivative of #sin^3x#? Question #eefeb Question #af14f See all questions in Limits Involving Trigonometric Functions Impact of this question 1379 views around the world You can reuse this answer Creative Commons License