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Find the values of \alpha and \beta such that   \lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x-1}-\alpha x-\beta\right)=0

 

Option: 1

(-1,-1)


Option: 2

(-1,1)


Option: 3

(1,-1)


Option: 4

(1,1)


Answers (1)

best_answer

Given

\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x-1}-\alpha x-\beta\right)=0

\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{x^2+1-\alpha x(x-1)-\beta(x-1)}{x-1}\right)=0 \\\\ & \lim _{x \rightarrow \infty}\left(\frac{x^2(1-\alpha)+x(\alpha-\beta)+(1+\beta)}{x-1}\right)=0 \end{aligned}

Since the limit of the given expression is zero, the degree of the numerator should be less than the degree of the denominator.

So,(1-\alpha)=0 \text { and }(\alpha-\beta)=0

 By solving the above two equations, the values of \alpha  and  \beta are obtained as(\alpha ,\beta) = (1,1)

Posted by

Irshad Anwar

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