If lim _{n rightarrow infty}left(sqrt{n^{2}-n | vedclass

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(C) We are given $\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-n-1}+n \alpha+\beta\right)=0$.For the limit to exist and be finite,the coefficient of

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$-4$

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(C) We are given $\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-n-1}+n \alpha+\beta\right)=0$.For the limit to exist and be finite,the coefficient of $n$ must be zero.$\sqrt{n^{2}-n-1} = n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} = n\left(1-\frac{1}{2}(\frac{1}{n}+\frac{1}{n^2}) - \frac{1}{8}(\frac{1}{n}+\frac{1}{n^2})^2 + \dots\right) = n - \frac{1}{2} - \frac{1}{2n} - \frac{1}{8n} + \dots = n - \frac{1}{2} + O(\frac{1}{n})$.Substituting this into the limit: $\lim _{n \rightarrow \infty} (n - \frac{1}{2} + n\alpha + \beta) = 0$.Grouping terms by powers of $n$: $\lim _{n \rightarrow \infty} (n(1+\alpha) + (\beta - \frac{1}{2})) = 0$.For this to be zero,we must have $1+\alpha = 0 \implies \alpha = -1$ and $\beta - \frac{1}{2} = 0 \implies \beta = \frac{1}{2}$.Thus,$8(\alpha+\beta) = 8(-1 + \frac{1}{2}) = 8(-\frac{1}{2}) = -4$.

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

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