Lim n →∞, sqrt(n2-n-1)+nα+β=0 then 8(α+β)

If $\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0 \text{ then } 8(\alpha + \beta )$ is equal to

(A) 4
(B) –8
(C) –4
(D) 8

Solution:

Tip for solving this question:

Simplify the limit

Find value of $\alpha \text{ and } \beta$

Step 1 of 3:

Given, $\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$

On rationalisation, we get

$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + (n\alpha + \beta )} \right)\left[ {\frac{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}}{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}}} \right] = 0$

$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\sqrt {{n^2} - n - 1} } \right)}^2} - {{(n\alpha + \beta )}^2}}}{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}} = 0 \{ \because (a + b)(a - b) = ({a^2} - {b^2})\}$

$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\sqrt {{n^2} - n - 1} } \right)}^2} - {{(n\alpha + \beta )}^2}}}{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}} = 0 \hfill \\ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{({n^2} - n - 1) - ({n^2}{\alpha ^2} + 2n\alpha \beta + {\beta ^2})}}{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}} \hfill \\ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}(1 - {\alpha ^2}) - n(1 + 2\alpha \beta ) - (1 + {\beta ^2})}}{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}} \hfill \\ \end{gathered}$

Step 2 of 3:

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

Here, $(1 - {\alpha ^2}) = 0$ $\Rightarrow \alpha = \pm 1$

$\therefore \alpha = - 1 \{ \because \alpha < 0\}$

Now,

$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{ - n(1 + 2\alpha \beta ) - (1 + {\beta ^2})}}{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}} \hfill \\ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{n( - 1 - 2\alpha \beta ) - (1 + {\beta ^2})}}{{\left( {\sqrt {{n^2} - n - 1} - (n\alpha + \beta )} \right)}} \hfill \\ \end{gathered}$

Here, $( - 1 - 2\alpha \beta ) = 0$

$\begin{gathered} \Rightarrow \left( { - 1 - 2( - 1)\beta } \right) = 0 \hfill \\ \Rightarrow \beta = \frac{1}{2} \hfill \\ \end{gathered}$

Step 3 of 3:

Now, $8(\alpha + \beta ) = 8\left( { - 1 + \frac{1}{2}} \right)$

$\begin{gathered} = 8\left( { - \frac{1}{2}} \right) \hfill \\ = - 4 \hfill \\ \end{gathered}$

Final Answer:

Hence, Option (C) is correct.

Comments

Khushi says

October 18, 2023 at 6:03 am

Why is negative value of alpha taken? Nothing as such is mentioned in the question.

Plumsé says

February 11, 2024 at 10:31 am

in step 2 how’d you deduce that alpha is -ve and 1-alpha^2 = 0, it’s nowhere mentioned in the ques.

Pranam Jena says

March 12, 2024 at 4:25 pm

because if alpha is positive then it is not indeterminate (infinite + infinite)form so alpha is taken as negative (infinite – infinite)