Question

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

• A. 4
• B. -8
• C. -4
• D. 8

Answer 1

Answer: (C)

$$\begin{gathered} {\lim }_{n\to \infty }n{\left(1-\frac{n+1}{n^2}\right)}^{\frac{1}{2}}+\alpha n+\beta =0 \\ {\lim }_{n\to \infty }n\{1-\frac{1}{2}\left(\frac{n+1}{n^2}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(\frac{n+1}{n^2}\right)}^2+\dots ..+\alpha n+\beta =0 \\ {\lim }_{n\to \infty }n-\frac{1}{2}+\frac{1}{n}+\dots .+n\alpha +\beta =0 \\ \alpha =-1,\beta =\frac{1}{2} \\ 8(\alpha +\beta )=-4 \end{gathered}$$