Question

If \(\displaystyle\lim _{n \rightarrow \infty}\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)=0 \) then \(8(\alpha+\beta)\) is equal to :

• A. 4
• B. $-8$
• C. -4
• D. 8

Answer 1

The Correct Option is C

To solve the problem, we need to find the value of \(8(\alpha + \beta)\) given the limit condition:

\[ \lim_{n \rightarrow \infty} \left(\sqrt{n^2 - n - 1} + n\alpha + \beta\right) = 0 \]

We start by simplifying \(\sqrt{n^2 - n - 1}\) when \(n\) is very large. Notice that:

\[ \sqrt{n^2 - n - 1} = \sqrt{n^2(1 - \frac{1}{n} - \frac{1}{n^2})} \]

This simplifies to:

\[ \sqrt{n^2(1 - \frac{1}{n} - \frac{1}{n^2})} \approx n\sqrt{1 - \frac{1}{n} - \frac{1}{n^2}} \]

Using the binomial approximation \(\sqrt{1 + x} \approx 1 + \frac{x}{2}\) for small \(x\), we get:

\[ \sqrt{1 - \frac{1}{n} - \frac{1}{n^2}} \approx 1 - \frac{1}{2n} - \frac{1}{2n^2} \]

Thus:

\[ \sqrt{n^2 - n - 1} \approx n \left(1 - \frac{1}{2n} - \frac{1}{2n^2}\right) = n - \frac{1}{2} - \frac{1}{2n} \]

Now substitute back into the limit equation:

\[ \lim_{n \to \infty} \left(n - \frac{1}{2} - \frac{1}{2n} + n\alpha + \beta \right) = 0 \]

Reorganize terms:

\[ \lim_{n \to \infty} \left(n(1 + \alpha) + \beta - \frac{1}{2} - \frac{1}{2n} \right) = 0 \]

For the limit to be zero, the coefficient of \(n\) must be zero and the constant terms must sum to zero for large \(n\). Therefore, we have:

1. \(1 + \alpha = 0 \Rightarrow \alpha = -1\)
2. \(\beta - \frac{1}{2} = 0 \Rightarrow \beta = \frac{1}{2}\)

Thus, the values are \(\alpha = -1\) and \(\beta = \frac{1}{2}\).

Calculate \(8(\alpha + \beta)\):

\[ 8(\alpha + \beta) = 8(-1 + \frac{1}{2}) = 8(-\frac{1}{2}) = -4 \]

Therefore, the value of \(8(\alpha + \beta)\) is \(-4\).