The correct options are
A −2
B 0
C √3+1
limx→−1(tan(x+1)+aex+1+ax+a−ax+1−a2sin−1(x+1))(x+1)(√x+2+1)x+1=1
⇒limx→−1(tan(x+1)+a(ex+1−1)+a(x+1)x+1−a2sin−1(x+1))√x+2+1=1
⇒limx→−1⎛⎜
⎜
⎜
⎜⎝tan(x+1)x+1+a(ex+1−1)x+1+a1−a2sin−1(x+1)x+1⎞⎟
⎟
⎟
⎟⎠√x+2+1=1
⇒(1+a+a1−a2)2=1
⇒(1+2a1−a2)=±1
⇒a2+2a=0 or a2−2a−2=0
⇒a=0,−2 or a=1±√3