If limx→∞ x2+1x+1−ax−b=0 ,then

# If $\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{{x}^{2}+1}{x+1}-ax-b\right\}=0$ ,then

1. A

$a=1,b=1$

2. B

$a=-1,b=1$

3. C

$a=1,b=-1$

4. D

$a=-1,b=-1$

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### Solution:

We have,

$\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{{x}^{2}+1}{x+1}-ax-b\right\}=0$

Since the limit of the given expression is zero. Therefore, degree of the polynomial in numerator must be less than that of denominator. As the denominator is a first degree polynomial. So, numerator must be a constant i.e., a zero degree polynomial

and $a+b=0⇒a=1$and

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