Question
If $$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right) = 4,$$ then-
A.
$$a=1,\,\,b=4$$
B.
$$a=1,\,\,b=-4$$
C.
$$a=2,\,\,b=-3$$
D.
$$a=2,\,\,b=3$$
Answer :
$$a=1,\,\,b=-4$$
Solution :
Given:
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right) = 4 \cr
& \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + x + 1 - a{x^2} - ax - bx - b}}{{x + 1}} = 4 \cr
& \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {1 - a} \right){x^2} + \left( {1 - a - b} \right)x + \left( {1 - b} \right)}}{{x + 1}} = 4 \cr} $$
For this limit to be finite $$1 - a = 0\,\,\,\, \Rightarrow a = 1$$
Then given limit reduces to
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{ - bx + \left( {1 - b} \right)}}{{x + 1}} = 4 \cr
& \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{ - b + \frac{{\left( {1 - b} \right)}}{x}}}{{1 + \frac{1}{x}}} = 4 \cr
& \Rightarrow - b = 4\,\,\,\,{\text{or}}\,\,\,b = - 4 \cr
& {\text{Hence}}\,\,\,a = 1,\,\,\,b = - 4 \cr} $$