Finding Constants in Rational Function Limit at Infinity

To find the constants $a$ and $b$ such that $\lim_{x \to \infty} \left( \frac{x^2 + x + 1}{x + 1} - ax - b \right) = 4$

we first combine the terms inside the limit over a common denominator:

$\lim_{x \to \infty} \left( \frac{(x^2 + x + 1) - (ax + b)(x + 1)}{x + 1} \right) = 4$

Expanding the product in the numerator: $(ax + b)(x + 1) = ax^2 + (a + b)x + b$

Substituting this back into the numerator: $\lim_{x \to \infty} \left( \frac{x^2 + x + 1 - ax^2 - (a + b)x - b}{x + 1} \right) = 4$

Grouping terms by powers of $x$: $\lim_{x \to \infty} \left( \frac{(1 - a)x^2 + (1 - a - b)x + (1 - b)}{x + 1} \right) = 4$

For the limit as $x \to \infty$ to be a finite non-zero constant ($4$), the degree of the numerator cannot be greater than the degree of the denominator. Since the denominator has degree $1$, the coefficient of $x^2$ in the numerator must be zero:

$1 - a = 0 \implies a = 1$

Now, substituting $a = 1$ back into the limit expression gives:

$\lim_{x \to \infty} \left( \frac{(1 - 1 - b)x + (1 - b)}{x + 1} \right) = 4$ $\lim_{x \to \infty} \left( \frac{-bx + (1 - b)}{x + 1} \right) = 4$

Dividing the numerator and the denominator by $x$:

$\lim_{x \to \infty} \left( \frac{-b + \frac{1 - b}{x}}{1 + \frac{1}{x}} \right) = -b$

Equating the result of the limit to the given value:

$-b = 4 \implies b = -4$

Therefore, the required values are $a = 1$ and $b = -4$.

This corresponds to Option (B).