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

Question

Evelyn Agard · Accepted Answer

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

Isabella Ackerman · Answer

undefined To find the limit of (1-cos(4x))/(1-cos(3x)) as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (-4sin(4x)) / (-3sin(3x)). Evaluating this expression as x approaches 0, we have (-4sin(0)) / (-3sin(0)), which simplifies to 0/0. Applying L'Hôpital's Rule again, we differentiate the numerator and denominator once more. This gives (-16cos(4x)) / (-9cos(3x)). Substituting x=0, we have (-16cos(0)) / (-9cos(0)), which simplifies to -16/-9. Therefore, the limit of (1-cos(4x))/(1-cos(3x)) as x approaches 0 is 16/9.