Question

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

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

Answer 1

Answer: (A), (C)

$b=1$

$\underset{n\to \infty }{lim}[an-\frac{1+n^2}{1+n}]=b\Rightarrow \underset{n\to \infty }{lim}\left[\frac{an+(a-1)n^2-1}{1+n}\right]=b$
For limit to be finite the numerator degree should be equal to the denominator degree.
Hence, the coefficient of $n^2$ should be equal to $0$
$\Rightarrow a-1=0\Rightarrow a=1$
Now,
$\underset{n\to \infty }{lim}\left[\frac{n-1}{1+n}\right]=b\Rightarrow b=1$