Question

If $\underset{n\to \infty }{\lim }\left(an-\frac{1+n^2}{1+n}\right)=b$, a finite number, then

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

Answer 1

Answer: (A), (C)

Required limit $=$
$\underset{n\to \infty }{\lim }\frac{an(1+n)-\left(1+n^2\right)}{1+n}$
$=\underset{n\to \infty }{\lim }\frac{(a-1)n^2+an-1}{1+n}=b$
(a finite number)
$\Rightarrow$ coeff. of $n^2$ must be $0$ as degree of deniminator is $1$ .
$\Rightarrow a=1$ putting this value,
$b=\underset{n\to \infty }{\lim }\frac{n-1}{1+n}$
$=\underset{n\to \infty }{\lim }\frac{1-\frac{1}{n}}{1+\frac{1}{n}}=1$
Hence, $a=1$ and $b=1$