Question

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

Answer 1

`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`