find-lim-n-n-p-sin-2-n-n-p-1-with0-lt-p-lt-1-

Question Number 34774 by abdo mathsup 649 cc last updated on 11/May/18

f i n d {l i m}_{n \to + \infty} \frac{n^{p} {s i n}^{2} (n!)}{n^{p + 1}} w i t h 0 < p < 1 .

Commented by abdo mathsup 649 cc last updated on 11/May/18

w e h a v e \frac{n^{p} {s i n}^{2} (n!)}{n^{p + 1}} = \frac{{s i n}^{2} (n!)}{n} b u t

0 ⩽ {s i n}^{2} (n!) ⩽ 1 \Rightarrow 0 ⩽ \frac{{s i n}^{2} (n!)}{n} ⩽ \frac{1}{n} \to 0 (n \to + \infty) s o

{l i m}_{n \to + \infty} \frac{n^{p} {s i n}^{2} (n!)}{n^{p + 1}} = 0 .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/May/18

= \frac{l i m}{n \to + \infty} \frac{{s i n}^{2} (n!)}{n}

l e t t = \frac{1}{n} w h e n n \to + \infty t \to + 0

\frac{l i m}{t \to + 0} t . {{s i n}^{2} (\frac{1}{t})!}

= 0 \times a n y v a l u e b e t w e e n + / - 1

= 0

=