Question

If $\underset{x\to \infty }{Lim}[\frac{x^2+1}{x+1}-ax-b]=b$, where a, b are constants, then the value of a+b, is

• A. $0$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $0$

Answer 1

The correct option is B $\frac{1}{2}$

$\underset{x\to \infty }{lim}[\frac{x^2+1}{x+1}-ax-b]=b\underset{x\to \infty }{lim}[\frac{x^2+1}{x+1}-ax]=2b\underset{x\to \infty }{lim}\left[\frac{x^2+1-ax(x+1)}{x+1}\right]=2b\underset{x\to \infty }{lim}\left[\frac{x^2(1-a)+(1-ax)}{x+1}\right]=2b\underset{x\to \infty }{lim}\frac{x^2\left[\right(1-a)+\left(\frac{1-ax}{x^2}\right)]}{x(1+\frac{1}{x})}=2b$
To remove the indeterminate form, a should be equal to 1 i.e., a =1
i.e., $\underset{x\to \infty }{lim}\frac{x\left(\frac{1-ax}{x^2}\right)}{(1+\frac{1}{x})}=2b\Rightarrow \underset{x\to \infty }{lim}\left(\frac{\frac{1}{x}-a}{1+\frac{1}{x}}\right)=-a=2bButa=1;\therefore b=-\frac{1}{2}\therefore a+b=1-\frac{1}{2}\Rightarrow a+b=\frac{1}{2}$.