Question

For $0<p<1$ the value of $\underset{n\to \infty }{lim}\frac{n^{p}si{n}^2(n!)}{n+1}$ , is equal to?

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

Answer 1

The correct option is A $0$

We have,
$\underset{n\to \infty }{lim}\frac{n^{p}si{n}^2(n!)}{n+1}=\underset{n\to \infty }{lim}\frac{si{n}^2(n!)}{1+\frac{1}{n}}\times n^{p-1}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{n^{p}si{n}^2(n!)}{n+1}=\underset{n\to \infty }{lim}\frac{si{n}^2(n!)}{n^{1-p}}\times \frac{1}{1+\frac{1}{n}}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{n^{p}si{n}^2(n!)}{n+1}=\frac{\text{An oscillating number}}{\infty }\times 1=0\times 1=0$.
Hence, option 'A' is correct.