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Q.
If $\displaystyle\lim _{n \rightarrow \infty}\left(a n-\frac{1+n^{2}}{1+n}\right)=b$, a finite number, then
Limits and Derivatives
Solution:
Required limit $=$
$\displaystyle\lim _{n \rightarrow \infty} \frac{a n(1+n)-\left(1+n^{2}\right)}{1+n}$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{(a-1) n^{2}+a n-1}{1+n}=b$
(a finite number)
$\Rightarrow $ coeff. of $n^{2}$ must be $0$ as degree of deniminator is $1$ .
$\Rightarrow a =1$ putting this value,
$b =\displaystyle\lim _{ n \rightarrow \infty} \frac{ n -1}{1+ n }$
$=\displaystyle\lim _{ n \rightarrow \infty} \frac{1-\frac{1}{ n }}{1+\frac{1}{ n }}=1$
Hence, $a=1$ and $b=1$