Question

If $lim_{ x \to \infty} \bigg( \frac{x^2 + x + 1}{x + 1} - ax - b \bigg)$ = 4, then

• A. a = 1, b = 4
• B. a = 1, b = - 4
• C. a = 2, b = - 3
• D. a = 2, b = 3

Answer 1

Answer: (B)

a = 1, b = - 4

Answer 2

PLAN $\bigg( \frac{\infty}{\infty}\bigg)$ form
lim_{ x \to \infty} \frac{a_0 x^n + a_1 x^{n - 1} + .... + a_n}{ b_0 x^m + b_1 x^{m - 1} + .... + b_m} = \Bigg \\{ \begin{array} \ 0, \\\ \frac{a_0}{b_0}. \\\ \+ \infty, \\\ \- \infty, \\\ \end{array} \begin{array} \ if n < m \\\ if n = m \\\ if n > m \ and \ a_0 b_0 > 0 \\\ if n > m \ and \ a_0 b_0 < 0 \\\ \end{array}
Description of Situation As to make degree of
numerator equal to degree of denominator.
$lim_{ x \to \infty} \bigg( \frac{x^2 + x + 1}{x + 1} - ax - b \bigg) = 4$
= $\Rightarrow lim_{ x \to \infty} \frac{x^2 + x + 1 - ax^2 - ax - bx - b}{ x + 1} = 4$
$\Rightarrow lim_{ x \to \infty} \frac{ x^2 (1 - a) + x ( 1 - a - b) + (1 - b)}{ x + 1} = 4$
Here, we make degree of numerator
\hspace12mm = degree of denominator
$\therefore 1 - a = 0 \Rightarrow a = 1$
and $lim_{x \to \infty} \frac{x ( 1 - a - b) + (1 - b)}{ x + 1} = 4$
$\Rightarrow 1 - a - b = 4$
$\Rightarrow b = - 4$ $[ \because (1 - a) = 0]$.