#lim_(x->0) ((1-sqrt(cos6x))/(x^2)) = # ? Calculus Differentiating Trigonometric Functions Limits Involving Trigonometric Functions 1 Answer Cesareo R. Sep 25, 2016 #9# Explanation: #lim_(x->0) ((1-sqrt(cos6x))/(x^2))# # (1-sqrt(cos6x))/(x^2) = (1-cos6x)/(x^2(1+sqrt(cos6x))# but #cos6x=1-2sin^2 3x# so # (1-cos6x)/(x^2(1+sqrt(cos6x))) = (2sin^2 3x)/(x^2(1+sqrt(cos6x))# Finally # (1-sqrt(cos6x))/(x^2) = 2(9(sin(3x)/(3x))^2)/(1+sqrt(cos6x))# then #lim_(x->0) ((1-sqrt(cos6x))/(x^2)) = lim_(x->0) 2(9(sin(3x)/(3x))^2)/(1+sqrt(cos6x)) = 9# 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 5346 views around the world You can reuse this answer Creative Commons License