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If $\lim\limits_{x\to \infty}\big(\large\frac{x^3+1}{x^2+1}$$-(ax+b)\big)=2$ then

$\begin{array}{1 1}(a)\;a=1,b=1&(b)\;b=1,b=2\\(c)\;a=1,b=-2&(d)\;None\;of\;these\end{array}$

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$\lim\limits_{x\to \infty}\bigg[\large\frac{x^3+1}{x^2+1}$$-(ax+b)\bigg]=2$
$\lim\limits_{x\to \infty}\large\frac{x^3(1-a)-bx^2-ax+(1-b)}{x^2+1}$$=2$
$\lim\limits_{x\to \infty}\large\frac{x(1-a)-b-\Large\frac{a}{x}+\frac{(1-b)}{x^2}}{1+\Large\frac{1}{x^2}}$$=2$
$\Rightarrow 1-a=0$
$-b=2$
$\therefore a=1,b=-2$
Hence (c) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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