Let α,β∈R be such that limx→0 x2sin⁡βxαx−sin⁡x=1. Then, 6(α+β) equals

We find thatlimx→0 x2sin⁡βxαx−sin⁡x=limx→0 x(sin⁡βx)α−sin⁡x=0α−1=0, if α≠1.But, it is given that limx→0 x2sin⁡βxαx−sin⁡x=1. So, α=1Now,limx→0 x2sin⁡βxαx−sin⁡xx=1⇒ limx→0 βx3sin⁡βxβxx−sin⁡x=1        [∵α=1]=limx→0 sin⁡βxβxx−sin⁡xx3=1β=6=1β⇒β=16                ∵limx→0 x−sin⁡xx3=16Hence, 6(α+β)=61+16=7.