Question

If $\underset{n\to \infty }{lim}(an-\frac{1+n^2}{1+n})=b$, where a and b are finite numbers, then

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

Answer 1

Answer: (A), (C)

$b=1$
$a=1$

$limit=\underset{n\to \infty }{lim}\frac{an(1+n)-(1+n^2)}{1+n}$

$=\underset{n\to \infty }{lim}\frac{(a-1)n^2+an-1}{n+1}$

$=\infty$ if $a-1\ne 0$
If $a-1=0$, limit $=\underset{n\to \infty }{lim}\frac{an-1}{n+1}=a=b$
$\therefore a=b=1$