If Ltn→∞(n2−n−1+nα+β)=0 then 8(α+β)=

Ltn→∞(n2−n−1+nα+β)=0 ⇒Ltn→∞(n(1-(n+1n2))12+nα+β=0 ⇒ Ltn→∞(n[1-12(n+1n2)+12(-12)2!(n+1n2)2+....]+nα+β=0 ⇒ Ltn→∞n-12+1n+....+nα+β=0 ⇒ Ltn→∞(α+1)n+(β-12)+1n+....=0 ⇒α+1=0 and β-12=0 ⇒α=-1,β=12 Now 8(α+β)=-4

If Ltn→∞(n2−n−1+nα+β)=0 then 8(α+β)=

Ltn→∞(n2−n−1+nα+β)=0 ⇒Ltn→∞(n(1-(n+1n2))12+nα+β=0 ⇒ Ltn→∞(n[1-12(n+1n2)+12(-12)2!(n+1n2)2+....]+nα+β=0 ⇒ Ltn→∞n-12+1n+....+nα+β=0 ⇒ Ltn→∞(α+1)n+(β-12)+1n+....=0 ⇒α+1=0 and β-12=0 ⇒α=-1,β=12 Now 8(α+β)=-4

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If $\underset{n \to \infty}{L t} (\sqrt{n^{2} - n - 1} + n α + β) = 0 t h e n 8 (α + β) =$

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-8

-4

answer is C.

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Detailed Solution

$\underset{n \to \infty}{L t} (\sqrt{n^{2} - n - 1} + n α + β) = 0 \Rightarrow \underset{n \to \infty}{L t} (n (1 - {(\frac{n + 1}{n^{2}}))}^{\frac{1}{2}} + n α + β = 0 \Rightarrow \underset{n \to \infty}{L t} (n[1- \frac{1}{2} (\frac{n + 1}{n^{2}})+ \frac{\frac{1}{2} (\frac{- 1}{2})}{2!} {(\frac{n + 1}{n^{2}})}^{2} + . . . .] + n α + β = 0 \Rightarrow \underset{n \to \infty}{L t} n- \frac{1}{2} + \frac{1}{n} +....+nα+β=0 \Rightarrow \underset{n \to \infty}{L t} (α+1)n+(β- \frac{1}{2})+ \frac{1}{n} +....=0 \Rightarrow α +1 = 0 a n d β - \frac{1}{2} = 0 \Rightarrow α = - 1, β = \frac{1}{2} N o w 8 (α + β) = - 4$