We have,
limx→0sin(a+x)−sin(a−x)−2sinaxsinx
=limx→02sin(a+x+a−x)2cos(a+x−a+x)2−2sinaxsinx
=limx→02sin2a2cos(2x)2−2sinaxsinx
=limx→02sinacosx−2sinaxsinx
=limx→02sina(cosx−1)xsinx
=limx→02sina(1−2sin2x2−1)xsinx
=limx→0−4sinasin2x2xsinx
=limx→0−4sinasin2x2x2sinx2cosx2
=limx→0−2sinasinx2xcosx200form
Applying L’ Hospital rule and we get,
=limx→0−2sinacosx2×(12)x(−sinx2)12+cosx2
=limx→0−sinacosx2x2(−sinx2)+cosx2
Taking limit and we get,
=−sinacos00(−sin0)+cos0
=−sina0+1
=−sina
Hence, this is the answer.