Question

If $\displaystyle\lim _{n \rightarrow \infty}\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)

$ \displaystyle\lim _{n \rightarrow \infty} n\left(1-\frac{n+1}{n^2}\right)^{\frac{1}{2}}+\alpha n+\beta=0$
$ \displaystyle\lim _{n \rightarrow \infty}\left\{11-\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+\ldots .\right\}+\alpha n+\beta=0 $
$ \displaystyle\lim _{n \rightarrow \infty} n-\frac{1}{2}+\frac{1}{n}+\ldots .+n \alpha+\beta=0$
$ \alpha=-1, \beta=\frac{1}{2}$
$ 8(\alpha+\beta)=-4$