Question

$\displaystyle\lim _{n \rightarrow \infty} \frac{n^{k} \sin ^{2}(n !)}{n+2} 0 < k < 1$, is equal to

• A. $\infty$
• B. 1
• C. 0
• D. None of these

Answer 1

Answer: (C)

$\displaystyle\lim _{n \rightarrow \infty} \frac{n^{k} \sin ^{2}(n !)}{n+2}$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{n^{k} \sin ^{2}(n !)}{n\left(1+\frac{2}{n}\right)}$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{\sin ^{2}(n !)}{n^{1-k}\left(1+\frac{2}{n}\right)}$
$=\frac{\text { a finite quantity }}{\infty}$
$\left[\because \sin ^{2}(n !)\right.$ always lies between $0$ and $1$.
Also, since $1-k>0, \therefore n^{1-k} \rightarrow \infty$ as $n \rightarrow \infty$ ]
$=0 .$