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

The Correct Option is B

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]$.