If $\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-n-1}+n \alpha+\beta\right)=0$ then $8(\alpha+\beta)$ is equal to :

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Limits

lim_{n \to \infty} (n^{2} - n - 1 + n α + β) = 0

then

8 (α + β)

is equal to :

Answer: (C)

Rewrite the limit:

n \to \infty lim n (1 - \frac{n + 1}{n ^{2}})^{\frac{1}{2}} + α n + β = 0

Using binomial expansion for large

n

(1 - \frac{n + 1}{n ^{2}})^{\frac{1}{2}} \approx 1 - \frac{1}{2} \cdot \frac{n + 1}{n ^{2}} = 1 - \frac{1}{2 n} - \frac{1}{2 n ^{2}}

Thus:

n (1 - \frac{1}{2 n} - \frac{1}{2 n ^{2}}) + α n + β = n - \frac{1}{2} - \frac{1}{2 n} + α n + β

For the limit to be zero as

n \to \infty

, the coefficient of

n

must vanish and the constant term must be zero:

1 + α = 0 ⟹ α = - 1

- \frac{1}{2} + β = 0 ⟹ β = \frac{1}{2}

Therefore:

8 (α + β) = 8 (- 1 + \frac{1}{2}) = 8 (- \frac{1}{2}) = - 4

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option

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