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Lim 1- cos(4x)/sin (2x) x>0(X tends to 0)
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its 0
yes, I see that but I am just wondering whether to just plug x=0 into the expression or is there an intermidiate step where we have to simplify or ...............
first consider the function [1- cos(4x)]/sin (2x)= [1-cos^2 2x +sin^2 2x]/sin2x =2sin^ 2x/sin 2x = 2 sin 2x now apply the limits, you will be getting 0
!!!!!!!!!!!!!GOT IT !!!!!!!!!!!!!!!!!
do u know lhopitals rule?
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