Let α and βbe the roots of ax2+bx+c=0,then Ltx→α1−cos(ax2+bx+c)(x−α)2=
a2(α−β)22
a22(α−β)2
a2(α−β)2
−a22(α−β)2
ax2+bx+c=a(x−α)(x−β)given limit is Limx→α2sin2(a(x−α)(x−β)2)2(x−α)2=a2(α−β)22