How do you find the limit of #(1-cos(4x))/(1-cos(3x)# as x approaches 0?

1 Answer
Cesareo R.
May 17, 2016

#lim_{x->0}(1 - cos(4 x))/(1 - cos(3 x)) = 16/9#

Explanation:

The series expansion of #cos(alpha x)# for #x = 0# in an open set containing #0# is given by#sum_{i=0}^{infty} (-1)^i((alpha x)^{2i})/(2n!)#

Then
#1-cos(4x) = 8 x^2 - (32 x^4)/3 + (256 x^6)/45+...#
also
#1-cos(3x) = (9 x^2)/2 - (27 x^4)/8 + (81 x^6)/80+...#
so
#1-cos(4x) = x^2(8 - (32 x^2)/3 + (256 x^4)/45+...)#
also
#1-cos(3x) = x^2(9/2 - (27 x^2)/8 + (81 x^4)/80+...)#

Substituting in the fraction
#(1 - cos(4 x))/(1 - cos(3 x)) = ((8 - (32 x^2)/3 + (256 x^4)/45+...))/((9/2 - (27 x^2)/8 + (81 x^4)/80+...))#
computing the limit for #x->0# we obtain
#lim_{x->0}(1 - cos(4 x))/(1 - cos(3 x)) = 16/9#

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