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  • <img alt="\mathrm{\lim _{n \rightarrow \infty} n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+n \alpha+\beta=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20%5Crightarro

\mathrm{\lim _{n \rightarrow \infty} n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+n \alpha+\beta=0}, find values of \mathrm{\alpha, \beta} and compute \mathrm{8(\alpha+\beta)}

Option: 1

2


Option: 2

-4


Option: 3

1


Option: 4

5


Answers (1)

best_answer

We have

\begin{aligned} \mathrm{n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+n \alpha+\beta} &\mathrm{ =\left(n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+n \alpha+\beta\right) \frac{n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-(n \alpha+\beta)}{n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-(n \alpha+\beta)} }\\ &\mathrm{ =\frac{n^2\left(1-\frac{1}{n}-\frac{1}{n^2}\right)-\left(n^2 \alpha^2+2 n \alpha \beta+\beta^2\right)}{n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-(n \alpha+\beta)} }\\ &\mathrm{ =\frac{n^2\left(1-\alpha^2\right)-n(1+2 \alpha \beta)-\left(1+\beta^2\right)}{n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-(n \alpha+\beta)} .} \end{aligned}
Hence, if we take the limit of the right-hand side, it would not exist, unless \mathrm{1-\alpha^2=0}. Hence, either \alpha=-1 or 1. But if  \alpha is not negative, the left-hand side also would not exist. So, \alpha=-1. Thus, the right-hand side becomes

\mathrm{\frac{-n(1-2 \beta)-\left(1+\beta^2\right)}{n \sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+n-\beta} .}

Hence, the limit gives \mathrm{\frac{2 \beta-1}{2}=0. } 

Thus, \mathrm{\beta=\frac{1}{2 }}, and hence \mathrm{8(\alpha+\beta)=-4}

Posted by

seema garhwal

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