(√n2−n−1+nα+β)=0 [ This limit will be zero when α<0 as when α>0 then overall limit will be ∞.] ⇒
lim
n→α
‌
(√n2−n−1+nα+β)(√n2−n−1−(nα+β))
√n2−n−1−(nα+β)
=0 ⇒
lim
n→α
‌
(n2−n−1)−(nα+β)2
√n2−n−1−(nα+β)
=0 ⇒
lim
n→α
‌
n2−n−1−n2α2−2nαβ−β2
√n2−n−1−(nα+β)
=0 ⇒
lim
n→α
‌
n2(1−α2)−n(1+2αβ)−(1+β2)
√n2−n−1−(nα+β)
Here power of " n " in the numerator is 2 and power of " n " in the denominator is 1. To get the value of limit equal to zero power of " n " should be equal in both numerator and denominator, otherwise value of limit will be infinite (∞). ∴ Coefficient of n2 should be 0 in this case. ∴1−α2=0 ⇒α=±1 But α should be <0 ∴α=+1‌ not possible ‌ ∴α=−1 ⇒
lim
n→α
‌
0−n(1+2αβ)−(1+β)
n[√1−‌
1
n
−‌
1
n2
−α−‌
β
n
]
=0 Divide numerator and denominator by n then we get, ⇒