Question

lf $\underset{x\to \infty }{lim}(\frac{x^2-1}{x+1}-ax-b)=2$ then

• A. $a=1$ and $b=-3$
• B. $a=1$ and $b=2$
• C. $a=0$ and $b=-1$
• D. $a=2$ and $b=-1$

Answer 1

Answer: (A)

$a=1$ and $b=-3$

$\underset{x\to \infty }{lim}(\frac{x^2-1}{x+1}-ax-b)=2$
$\Rightarrow \underset{x\to \infty }{lim}(\frac{(x-1)(x+1)}{x+1}-ax-b)=2$
$\Rightarrow \underset{x\to \infty }{lim}(x-1-ax-b)=2$
For above limit to exist $a=1$
$\Rightarrow \underset{x\to \infty }{lim}(-1-b)=2$
$\Rightarrow b+1=-2\Rightarrow b=-3$