Vikas TU
Last Activity: 8 Years ago
Since α,β are the roots of ax2+bx+c=0
limx→α (1−cos(x−α)(x−β)) / (x−α)2
⇒limx=α 2sin2(x−α)(x−β)2a2(x−α)2(x−β)24×a2(x−α)2(x−β)24(x−α)2
⇒limx=α {(2sin2(x−α)(x−β) / 2) / ( a2(x−α)2(x−β)2 / 4) } × {a2(x−α)2(x−β)2 / 4(x−α)2 }
⇒2×a2(α−β)2 / 4
⇒a2(α−β)2 /2