Given, x→0limαx(e3x−1)αx−(e3x−1)=β
⇒x→0lim3xαx(e3x−1)⋅3xαx+1−e3x=β
⇒x→0lim3αx2αx+1−e3x=β as (x→0limxex−1=1)
⇒x→0lim6αxα−3e3x=β [Using L' Hospital rule as (00form)]
Now for limit to exist, α=3
So, x→0lim6×3×x3−3e3x=β
So, β=x→0lim6x−(e3x−1)=x→0lim6−3e3x=−21
∴α+β=25