Lim x->o (2-cosx-cos3x)/(1-cos3x) without the use l hopital's Rule?

Let u = cos x
cos 3x = 4*cos^3 x - 3*cos x = 4u^3 - 3u

As x → 0 , so u → 1

( 2 - cos x - cos 3x ) / ( 1 - cos 3x )
= ( 2 - u - 4u^3 + 3u ) / ( 1 - 4u^3 + 3u )
= ( - 4u^3 + 2u + 2 ) / ( - 4u^3 + 3u + 1 )
= - 2( 2u^3 - u - 1 ) / [ - ( 4u^3 - 3u - 1 ) ]
= 2( u - 1 )( 2u^2 + 2u + 1 ) / [ ( u - 1 )( 4u^2 + 4u + 1 ) ]
→ 2( 2u^2 + 2u + 1 ) / ( 4u^2 + 4u + 1 ) , because u → 1 implies u ≠ 1
→ 2( 2 + 2 + 1 ) / ( 4 + 4 + 1 )
= 10 / 9 ..... Ans