Question

What is the limit as `x` approaches `0` of `(4x^2) / (1- cos2x)`?

Answer 1

`lim_{x->0} (4x^2)/(1-cos(2x))=2`

Explanation:

First we will use
`cos(a+b)=cos(a)cos(b)-sin(a)sin(b)`. If `a=b` then
`cos(2a)=1-2sin^2(a)`. Minding this result
`(4x^2)/(1-cos(2x)) = (4x^2)/(2sin^2(x)) = 2(x/(sin(x)))^2`
but `lim_{x->0}(x/(sin(x)))=1`
so
but `lim_{x->0} (4x^2)/(1-cos(2x))=2`