Question

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

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

Answer 1

Answer: (C)

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