If _ n → ∞ ( n ^ 2 - n - 1 + nα + β) = 0 then 8(α + β) is equal to

Given, $\lim_{n \to \infty}\sqrt{{n}^{2} - n - 1} + \alphan + \beta = 0$

Now rewriting the expression we get,

$\lim_{n \to \infty}\left({n}^{2} - n - 1\right)\dfrac{1}{2} + \alphan = - \beta$

$⇒\lim_{n \to \infty}n\left(1 - \dfrac{1}{n} - \dfrac{1}{{n}^{2}}\right)\dfrac{1}{2} + \alphan = - \beta$

Now on using binomial approximation ${\left(1 + x\right)}^{n} = 1 + nx$ we get,

$\lim_{n \to \infty}n\left(1 - \dfrac{1}{2n} - \dfrac{1}{2{n}^{2}}\right) + \alphan = - \beta$

$⇒ \lim_{n \to \infty}n - \dfrac{1}{2} - \dfrac{1}{2n} + \alphan = - \beta$

$⇒ \lim_{n \to \infty}n\left(1 + \alpha\right) - \dfrac{1}{2} - \dfrac{1}{2n} = - \beta$

Now for limit to exist \left(1 + \alpha\right) = 0 ⇒ \alpha = - 1$</p><p>Now on putting the value of$\alpha$we get,</p><p>$\lim_{n \to \infty}n\left(1 + \alpha\right) - \dfrac{1}{2} - \dfrac{1}{2n} = - \beta$</p><p>$⇒ \lim_{n \to \infty} - \dfrac{1}{2} - \dfrac{1}{2n} = - \beta$</p><p>$⇒ - \dfrac{1}{2} - 0 = - \beta$</p><p>$⇒ \beta = \dfrac{1}{2}$</p><p>So,$8\left(\alpha + \beta\right) = 8\left(- 1 + \dfrac{1}{2}\right) = 8 \times \dfrac{- 1}{2} = - 4