Two possible answers for α and β

Question

So there was this limits question in the JEE Main 2022 exam:

If $\lim\limits_{n \to \infty}( \sqrt{n^{2}-n-1}+n\alpha +\beta)=0$

Find $8(\alpha + \beta)$

It had the following options:

A. 4
B. -8
C. 8
D. -4

I tried to solve it by multiplying the numerator and denominator of the L.H.S. with the conjugate of the numerator and got the following expression:

$\lim\limits_{n \to \infty} (\frac{n^{2}(1-\alpha^{2})-n(1+2\alpha\beta)-(1+\beta^{2})}{\sqrt{n^{2}-n-1}-(n\alpha+\beta)})$

Now since $n$ is tending to infinity and the limit is equal to zero the power of $n$ will be more in the denominator than in the numerator, hence we can equate the coefficients of $n^{2}$ and $n$ in the numerator to zero which gives us $1=\alpha^{2}$ and $-1=2\alpha\beta$ so there are two possible $(\alpha,\beta)$: $(1,\frac{-1}{2})$ and $(-1,\frac{1}{2})$ for which there are two possible answers to $8(\alpha+\beta):4$ and $-4$

But there is only one correct answer to the question: $-4$

What is wrong with my reasoning and solution?

Answer 1

None of the other answers actually addresses why your method does not seem to work. I will show you how your method can be saved.

Firstly, you find out that the limit is equal to $\lim_{n\to\infty}\frac{a_n}{b_n}$ with $a_n:=n^{2}(1-\alpha^{2})-n(1+2\alpha\beta)-(1+\beta^{2})$ and $b_n:=\sqrt{n^{2}-n-1}-(n\alpha+\beta)$. This is correct!

Now, you want $a_n$ to be asymptotically weaker than $b_n$, which actually requires $a_n$ to be a constant, that is,

$$1-\alpha^2=0,\qquad 1+2\alpha\beta=0.$$

However, notice that this condition is not enough: If $b_n$ tends to a constant as well, then $\frac{a_n}{b_n}$ tends towards a non-zero constant, and your limit is not zero. Thus, we need to add the condition that $\alpha\neq 1$.

In total, we end up with a system $$1-\alpha^2=0,\qquad 1+2\alpha\beta=0,\qquad\alpha\neq 1,$$

which means that $\alpha=-1$ and $\beta=\frac 12$.

Answer 2

I assume $n\to+\infty$ so $$ \sqrt{n^2 - n - 1} = n\sqrt{1 - \frac1n - \frac1{n^2}} $$ therefore $$ \sqrt{n^2 - n - 1} + \alpha n +\beta =n\left(\sqrt{1 - \frac1n - \frac1{n^2}} +\alpha\right)+\beta $$ therefore necessarily one has $$ \alpha = -1 $$ in fact any value of $\alpha$ different from $-1$ makes the limit $\infty$. Now \begin{align*} n\left(\sqrt{1 - \frac1n - \frac1{n^2}} -1\right) &=n\frac{\left(\sqrt{1 - \frac1n - \frac1{n^2}} -1\right)\left(\sqrt{1 - \frac1n - \frac1{n^2}} +1\right)}{\sqrt{1 - \frac1n - \frac1{n^2}} +1}\\ &=n\frac{1 - \frac1n - \frac1{n^2}-1}{\sqrt{1 - \frac1n - \frac1{n^2}} +1}\to-\frac12 \end{align*} hence $\beta=\frac12$ which leads to $8(\alpha + \beta)=-4$.