Question

The value of the constant $\alpha$ and $\beta$ such that $\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-\alpha x-\beta )=0$ are respectively.

• A. $(1,1)$
• B. $(-1,1)$
• C. $(1,-1)$
• D. $(0,1)$

Answer 1

Answer: (C)

$(1,-1)$

$\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-\alpha x-\beta )=0$

$=\underset{x\to \infty }{lim}\left(\frac{(x^2+1)-(\alpha x+\beta )(x+1)}{(x+1)}\right)=0$

$\Rightarrow \underset{x\to \infty }{lim}\frac{x^2+1-\alpha x^2-\alpha x-\beta x-\beta }{x+1}=0$

$\Rightarrow \underset{x\to \infty }{lim}\frac{x^2(1-\alpha )+x(-\alpha -\beta )+1-\beta }{x+1}=0$

$\therefore 1-\alpha =0$ $\Rightarrow \alpha =1$

$\therefore -\alpha -\beta =0$

$\Rightarrow \beta =-\alpha$

$\therefore \beta =-1$

$(\alpha ,\beta )=(1,-1)$