If lim _{x rightarrow infty}left{frac{x^3+1}{ | vedclass

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(D) Given that $\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}=2$. Simplifying the expression: $\lim _{x \rightarrow

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$(1, -2)$

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(D) Given that $\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}=2$. Simplifying the expression: $\lim _{x \rightarrow \infty} \frac{x^3+1-(\alpha x+\beta)(x^2+1)}{x^2+1} = 2$ $\lim _{x \rightarrow \infty} \frac{x^3+1-(\alpha x^3+\alpha x+\beta x^2+\beta)}{x^2+1} = 2$ $\lim _{x \rightarrow \infty} \frac{(1-\alpha)x^3-\beta x^2-\alpha x+(1-\beta)}{x^2+1} = 2$. For the limit to exist and be finite,the coefficient of $x^3$ must be $0$: $1-\alpha = 0 \Rightarrow \alpha = 1$. Now the limit becomes: $\lim _{x \rightarrow \infty} \frac{-\beta x^2-x+(1-\beta)}{x^2+1} = 2$. Dividing numerator and denominator by $x^2$: $\lim _{x \rightarrow \infty} \frac{-\beta - \frac{1}{x} + \frac{1-\beta}{x^2}}{1 + \frac{1}{x^2}} = 2$. As $x \rightarrow \infty$,the terms with $x$ in the denominator approach $0$: $-\beta = 2 \Rightarrow \beta = -2$. Thus,the ordered pair is $(\alpha, \beta) = (1, -2)$.

If $\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}$ exists and is equal to $2$,then the ordered pair $(\alpha, \beta)$ of real numbers is

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