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

<p>$$\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$$</p> <p>[ This limit will be zero when $\alpha$ < 0 as when $\alpha$ > 0 then overall limit will be $\infty$. ]</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right)\left( {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$$</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$$</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2} - n - 1 - {n^2}{\alpha ^2} - 2n\alpha\beta - {\beta ^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$$</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2}\left( {1 - {\alpha ^2}} \right) - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + {\beta ^2}} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}$$</p> <p>Here power of "n" in the numerator is 2 and power of "n" in the denominator is 1.</p> <p>To get the value of limit equal to zero power of "n" should be equal in both numerator and denominator, otherwise value of limit will be infinite ($\infty$).</p> <p>$\therefore$ Coefficient of n² should be 0 in this case.</p> <p>$\therefore$ $1 - {\alpha ^2} = 0$</p> <p>$\Rightarrow \alpha = \, \pm \,1$</p> <p>But $\alpha$ should be < 0</p> <p>$\therefore$ $\alpha = \, + \,1$ not possible</p> <p>$\therefore$ $\alpha = - 1$.</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{0 - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + \beta } \right)} \over {n\left[ {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}} \right]}} = 0$$</p> <p>Divide numerator and denominator by n then we get,</p> <p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{ - \left( {1 + 2\alpha \beta } \right) - {{(1 + \beta )} \over n}} \over {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}}} = 0$$</p> <p>$$ \Rightarrow {{ - \left( {1 + 2\alpha \beta } \right) - 0} \over {\sqrt {1 - 0 - 0} - \alpha - 0}} = 0$$</p> <p>$\Rightarrow {{ - \left( {1 + 2\alpha \beta } \right)} \over {1 - \alpha }} = 0$</p> <p>$\Rightarrow - \left( {1 + 2\alpha \beta } \right) = 0$</p> <p>$\Rightarrow 1 + 2\alpha \beta = 0$</p> <p>$\Rightarrow 2\alpha \beta = - 1$</p> <p>$$ \Rightarrow \beta = - {1 \over {2\alpha }} = - {1 \over {2( - 1)}} = {1 \over 2}$$</p> <p>$\therefore$ $8\left( {\alpha + \beta } \right)$</p> <p>$= 8\left( { - 1 + {1 \over 2}} \right)$</p> <p>$= 8 \times - {1 \over 2}$</p> <p>$= - 4$</p>