Explanation

(c)
Given, $\lim_{x \rightarrow \infty}\left( \frac{x^{2} + 1}{x + 1} - \alpha x - \beta \right) = 0$
$\Rightarrow \ \ \ \ \ \ \ \lim_{x \rightarrow \infty}\left( \frac{x^{2} + 1 - \alpha(x^{2} + x) - \beta(x + 1)}{x + 1} \right) = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \lim_{x \rightarrow \infty}\left( \frac{2x - \alpha(2x + 1) - \beta(1)}{1} \right) = 0$
[by L’ Hospital’s rule]
If this limit is zero, then the function
$2x - \alpha(2x + 1) - \beta = 0$
or $x(2 - 2\alpha) - (\alpha + \beta) = 0$
Equating the coefficient of $x$ and constant terms, we get
$2 - 2\alpha = 0\$ and $\alpha + \beta = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \alpha = 1,\ \ \ \ \ \ \ \ \ \beta = - 1$