Question

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

• A. 0
• B. $\infty$
• C. 1
• D. None

Answer 1

Answer: (A)

limit $=\underset{n\to \infty }{\lim }\frac{{\sin }^2(n!)}{n^{1-p}(1+1/n)}(0<p<1)$
$=\frac{\text{ some real number in }[0,1]}{\infty }=0.(\because 1-p>0)$