$\lim_{n\rightarrow \infty} \ \ \ n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} +n\alpha + \beta=0$, find values of $\alpha,\beta$ and compute $$8(\alpha+\beta)$$
$$\lim_{n\rightarrow \infty}n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} +n\alpha+\beta=0$$
And since $n \rightarrow \infty$, the $\frac{1}{n^t}, (t=1,2..)$ terms will get cancelled so:
$n+n\alpha=0$ should be 0 for the limit to be defined
$$\implies \alpha=-1$$
And since the entire limit is $0$, and only $\beta$ is left so:
$\beta=0$
$$8(\alpha+\beta)=-8$$
The answer given in my textbook is $-4$. Please help me out with the mistake.
Best Answer
We have \begin{align} n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} +n\alpha + \beta &= \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)}\\ &=\frac{n^2\left(1-\frac{1}{n}-\frac{1}{n^2}\right) -(n^2\alpha^2 + 2n\alpha\beta+\beta^2)}{n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} -(n\alpha + \beta)}\\ &=\frac{n^2(1-\alpha^2)-n(1+2\alpha\beta)-(1+\beta^2)}{n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} -(n\alpha + \beta)}.\\ \end{align}
Hence, if we take the limit of the right-hand side, it would not exist, unless $1-\alpha^2= 0$. 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
$$\frac{-n(1-2\beta)-(1+\beta^2)}{n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} +n - \beta}.$$
Hence, the limit gives $\frac{2\beta-1}{2}=0$. Thus, $\beta=1/2$, and hence $8(\alpha+\beta)=-4$.
EDIT: To answer why yours is faulty, it's the part where you said $n+n\alpha$ must be zero for the limit to exist. What's right to say is only that $\alpha$ must be negative.