Question 357 of 401MediumMCQ
If \[\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0\], then \[8(\alpha+\beta)\] is equal to :
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Official solution
\[\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0\]
[ This limit will be zero when \[\alpha\] 0 then overall limit will be \[\infty \]. ]
\[ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right)\left( {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0\]
\[ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0\]
\[ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2} - n - 1 - {n^2}{\alpha ^2} - 2n\alpha\beta - {\beta ^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0\]
\[ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2}\left( {1 - {\alpha ^2}} \right) - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + {\beta ^2}} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}\]
Here power of "n" in the numerator is 2 and power of "n" in the denominator is 1.
To get the value of limit equal to zero power of "n" should be equal in both numerator and denominator, otherwise value of limit will be infinite (\[\infty \]).
\[\therefore\] Coefficient of n2 should be 0 in this case.
\[\therefore\] \[1 - {\alpha ^2} = 0\]
\[ \Rightarrow \alpha = \, \pm \,1\]
But \[\alpha\] should be
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