limx→0(1−cos2x)(3+cosx)xtan4x is equal to

limx→0(1−cos2x)(3+cosx)xtan4x
We know that 1−cos2x=2sin²x and limx→0tanx/x=1
So we have
limx→0(2sin²x)(3+cosx)x(4x)
=limx→0(2sin²x)(3+cosx)4x²/x
=limx→0(2sin²x/x²)(3+cosx)(4x)
=limx→0(2sin²x/x²)(limx→0(3+cosx))(limx→04x)
=2(1)(3+1)(0)
=0
Let's use L'Hopital's rule:
limx→0(1−cos2x)(3+cosx)xtan4x =limx→0(1−cos2x)(3+cosx)x(sin4x/cos4x)
=limx→0(1−cos2x)(3+cosx)xsin4x/cos4x
=limx→0(2sin²x)(3+cosx)x(4x)
=limx→0(8xsin²x)(3+cosx)/x
=limx→08xsin²x/x²(3+cosx)x
=limx→08sin²x/x(3+cosx)x
=limx→08sin²x/x²(3+cosx)
=8(1)(4)=32