Question

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

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

Answer 1

The correct option is C $b=1$

$\underset{n\to \infty }{lim}[an-\frac{1+n^2}{1+n}]=b$
$\Rightarrow \underset{n\to \infty }{lim}\left[\frac{an(1+n)-(1+n^2)}{1+n}\right]=b$
$\Rightarrow \underset{n\to \infty }{lim}\left[\frac{(a-1)n^2+an-1}{1+n}\right]=b$
$\Rightarrow \underset{n\to \infty }{lim}\left[\frac{(a-1)n+a-1/n}{1/n+1}\right]=b$
For above limit to exist, coefficient of $n$, should be zero
$\Rightarrow a=1$
so the limit of L.H.S. is $=a=1=b$ (given)