Mathematics - Limits Question with Solution | TestHub

Given, $\underset{n\to \infty }{\lim }\sqrt{n^2-n-1}+\alpha n+\beta =0$

Now rewriting the expression we get,

$\underset{n\to \infty }{\lim }{\left(n^2-n-1\right)}^{\frac{1}{2}}+\alpha n=-\beta$

$\Rightarrow$$\underset{n\to \infty }{\lim } n{\left(1-\frac{1}{n}-\frac{1}{n^2}\right)}^{\frac{1}{2}}+\alpha n=-\beta$

Now on using binomial approximation ${\left(1+x\right)}^n=1+nx$ we get,

$\underset{n\to \infty }{\lim } n\left(1-\frac{1}{2n}-\frac{1}{2n^2}\right)+\alpha n=-\beta$

$\Rightarrow \underset{n\to \infty }{\lim } n-\frac{1}{2}-\frac{1}{2n}+\alpha n=-\beta$

$\Rightarrow \underset{n\to \infty }{\lim } n\left(1+\alpha \right)-\frac{1}{2}-\frac{1}{2n}=-\beta$

Now for limit to exist $\left(1+\alpha \right)=0\Rightarrow \alpha =-1$

Now on putting the value of $\alpha$ we get,

$\underset{n\to \infty }{\lim } n\left(1+\alpha \right)-\frac{1}{2}-\frac{1}{2n}=-\beta$

$\Rightarrow \underset{n\to \infty }{\lim } -\frac{1}{2}-\frac{1}{2n}=-\beta$

$\Rightarrow -\frac{1}{2}-0=-\beta$

$\Rightarrow \beta =\frac{1}{2}$

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