Given, p(x)=0
⇒x2−x−2=0
⇒(x−2)(x+1)=0
⇒x=2,−1∴α=2
Now,x→2+limx−21−cos(x2−x−2)
=x→2+limx−22sin2(2x2−x−2)
=x→2+lim2⋅x−2∣sin(2x2−x−2)∣
=x→2+lim2⋅x−2sin(2x2−x−2) (because x>2)
=x→2+lim(2x2−x−2)⋅22sin(2x2−x−2)⋅(x−2x2−x−2)
=x→2+lim21⋅(x−2)(x−2)(x+1)
=23.