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

We need to find $$ 8(\alpha + \beta) $$ given that $$ \lim_{n \to \infty} \left(\sqrt{n^2 - n - 1} + n\alpha + \beta\right) = 0 $$.

As $$ n \to \infty $$, $$ \sqrt{n^2 - n - 1} \approx n\sqrt{1 - \frac{1}{n} - \frac{1}{n^2}} \approx n $$.

For the limit to be 0 (and not $$ \pm\infty $$), we need the coefficient of $$ n $$ in the expression to vanish, which requires $$ \alpha = -1 $$.

$$\sqrt{n^2 - n - 1} - n + \beta$$

Rationalize:

$$\sqrt{n^2 - n - 1} - n = \frac{(n^2 - n - 1) - n^2}{\sqrt{n^2 - n - 1} + n} = \frac{-n - 1}{\sqrt{n^2 - n - 1} + n}$$

$$\frac{-n - 1}{\sqrt{n^2 - n - 1} + n} = \frac{-n(1 + \frac{1}{n})}{n(\sqrt{1 - \frac{1}{n} - \frac{1}{n^2}} + 1)}$$

As $$ n \to \infty $$:

$$\to \frac{-1}{1 + 1} = -\frac{1}{2}$$

$$-\frac{1}{2} + \beta = 0 \implies \beta = \frac{1}{2}$$

$$8(\alpha + \beta) = 8\left(-1 + \frac{1}{2}\right) = 8 \times \left(-\frac{1}{2}\right) = -4$$

Therefore, $$ 8(\alpha + \beta) = -4 $$, which corresponds to Option C.