If lim _(n → ∞) ( √(n² - n - 1) + nα + β ) = 0, then 8(α+β) is equal to :

Solution

Maths sir · 4 steps

First kill the n term so the limit stays finite, then rationalise to find what β must cancel.

1. Fix alpha. For large n, √(n²-n-1) behaves like n. If 1+α ≠ 0, the expression grows like (1+α)n and cannot tend to 0. So α = -1.
2. Rationalise. With α = -1 we have an ∞ - ∞ form. Multiply and divide by √(n²-n-1)+n, then divide the top and bottom by n.

   √(n²-n-1)-n = (-n-1)/(√(n²-n-1)+n) = (-1-1/n)/(√(1-1/n-1/n²)+1) → -1/2

3. Fix beta. The whole limit is -1/2 + β = 0, so β = 1/2.
4. So. α + β = -1 + 1/2 = -1/2, and 8(α+β) = -4. The answer is (c).

The idea to keep. For a limit like √(an²+bn+c) + pn + q to be finite, first make the n terms cancel. Then rationalise: the leftover constant is b/2√a.

Answer from the NTA final answer key. Solution by PrepWiser.