Lim x tends to 0 (- sin x - 1/(1+x) + 1) / 2 sinx cos x ....0/0 case L'Hospital's rule applied
But it is still in 0/0 form .. So again differentiating
Lim x tends to 0 ( - cos x + 1/(1+x)^2 ) / 2 cos 2x .. 2 sin x cos x = sin 2x
( Lim x tends to 0 - cos x / 2 cos 2x ) + ( Lim x tends to 0 1/ 2 (1+x)^2 ) cos 2x )
Applying limits
= -1/2 + 1/2
= 0