Question

The values of constants a and b so that $\lim_{x \rightarrow \infty}\left( \frac{x^{2} + 1}{x + 1} - ax - b \right) = 0$ is

• A. $a = 0,b = 0$
• B. $a = 1,b = - 1$
• C. $a = - 1,b = 1$
• D. $a = 2,b = - 1$

Answer 1

Answer: (B)

$a = 1,b = - 1$

Answer 2

We have $\lim_{x \rightarrow \infty}\left( \frac{x^{2} + 1}{x + 1} - ax - b \right) = 0$

$\Rightarrow$ $\lim_{x \rightarrow \infty}\frac{x^{2}(1 - a) - x(a + b) + 1 - b}{x + 1} = 0$

Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than that of denominator. As the denominator is a first degree polynomial. So, numerator must be a constant i.e., a zero degree polynomial. ∴ $1 - a$ =0 and $a + b = 0$ ⇒ a = 1 and

b = –1. Hence, a = 1 and b = – 1.