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If $\lim\limits_{x\to \infty}\bigg[\large\frac{x^2+x+1}{x+1}$$-ax-b\bigg]=4$ then

$\begin{array}{1 1}(a)\;a=1,b=4&(b)\;a=1,b=-4\\(c)\;a=2,b=-3&(d)\;a=2,b=3\end{array}$

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Given :
$\lim\limits_{x\to \infty}\big(\large\frac{x^2+x+1}{x+1}-$$ax-b\big)=4$
$\lim\limits_{x\to \infty}\large\frac{x^2+x+1-ax^2-ax-bx-b}{x+1}$$=4$
$\lim\limits_{x\to \infty}\large\frac{(1-a)x^2+(1-a-b)x+(1-b)}{x+1}$$=4$
For the limit to be finite $1-a=0$
$\Rightarrow a=1$
Then the given limit reduces to
$\lim\limits_{x\to \infty}\large\frac{-bx+(1-b)}{x+1}$$=4$
$\Rightarrow \lim\limits_{x\to \infty}\large\frac{-b+\big(\Large\frac{1-b}{x})}{1+1/x}$$=4$
$\Rightarrow -b=4$ or $b=-4$
Hence $a=1,b=-4$
Hence (b) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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