MCQ
The value of the constant $\alpha $ and $\beta $ such that $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + 1}}{{x + 1}} - \alpha x - \beta } \right) = 0$ are respectively
  • A
    $(1, 1)$
  • B
    $(-1, 1)$
  • $(1, -1)$
  • D
    $(0, 1)$

Answer

Correct option: C.
$(1, -1)$
c
(c) $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + 1}}{{x + 1}} - 2x - \beta } \right) = 0$

==> $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}(1 - \alpha ) - x(\alpha + \beta ) + 1 - b}}{{x + 1}} = 0$

Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than denominator.

$\therefore$ $1 - \alpha = 0$ and $\alpha + \beta = 0$

==> $\alpha = 1$ and $\beta = - 1$.

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