Question

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

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

Answer 1

Answer: (C)

$(1,-1)$

Simplifying the above expression gives us
$\underset{x\to \infty }{lim}\left(\frac{x^2+1-\alpha x(x+1)-\beta (x+1)}{x+1}\right)$
$=\underset{x\to \infty }{lim}\left(\frac{x^2(1-\alpha )-x(\alpha +\beta )+1-\beta )}{x+1}\right)$
$=0$ implies that $Numerator<Denominator$.
Hence coefficient of $x^2$ will be 0. Therefore $\alpha =1$.
Hence
$\underset{x\to \infty }{lim}\left(\frac{-x(1+\beta )+1-\beta )}{x+1}\right)=0$
Applying L'Hopital's Rule we get
$\underset{x\to \infty }{lim}\left(\frac{-(1+\beta )}{1}\right)=0$
Or
$\beta +1=0$
Or
$\beta =-1$.
Hence
$(\alpha ,\beta )=1,-1$