Limits, Continuity and Differentiability Part 2

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