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

Question

$\lim_{n\rightarrow \infty} \ \ \ n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} +n\alpha + \beta=0$, 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.

Answer 1

The expression $\sqrt{1+x}$ can be written as the following series expansion

$$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \ ... \tag{1}$$

when $|x| < 1$.

Applying this to your limit with $x := - 1/n - 1/n^2, \ $ gives

$$\lim_{n \to \infty}n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}} + \alpha n + \beta \tag{2}$$

$$= \lim_{n \to \infty} n\left(1 + \frac{(-1/n - 1/n^2)}{2} - \ ...\right) + \alpha n + \beta \tag{3}$$

$$= \lim_{n \to \infty} n - \frac{1}{2} + \ ... \ + \alpha n + \beta. \tag{4}$$

For this limit to equal $0$, we need both the terms in $n$ and the constant terms to cancel.

Thus, it's clear that

$$\alpha = -1 \ \text{ and } \ \beta = \frac{1}{2} \tag{5}$$

Hence,

$$8(\alpha + \beta) = 8\left(-1 + \frac{1}{2}\right) = -4 \tag{6}$$

Keep in mind that when writing out the expansion in the limit, we neglected any terms of size $1/n^2$ or smaller, as they approach zero even when multiplied by $n$.

Answer 2

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.