The values of the constants alpha and beta su | vedclass

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(C) Given $\lim_{x 	o \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta ight) = 0$. Simplify the expression inside the limit: $\frac{x^2 + 1

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$(1, -1)$

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(C) Given $\lim_{x 	o \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta ight) = 0$. Simplify the expression inside the limit: $\frac{x^2 + 1 - (\alpha x + \beta)(x + 1)}{x + 1} = \frac{x^2 + 1 - (\alpha x^2 + \alpha x + \beta x + \beta)}{x + 1} = \frac{x^2(1 - \alpha) - x(\alpha + \beta) + (1 - \beta)}{x + 1}$. For the limit to be $0$ as $x 	o \infty$,the degree of the numerator must be less than the degree of the denominator. Therefore,the coefficient of $x^2$ must be $0$: $1 - \alpha = 0 \implies \alpha = 1$. Next,the coefficient of $x$ must also be $0$: $-(\alpha + \beta) = 0 \implies \alpha + \beta = 0$. Substituting $\alpha = 1$: $1 + \beta = 0 \implies \beta = -1$. Thus,the values are $\alpha = 1$ and $\beta = -1$.

The values of the constants $\alpha$ and $\beta$ such that $\lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta \right) = 0$ are respectively:

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