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

$$find{lim}_{n\to +\mathrm{\infty }}\frac{n^p{sin}^2(n!)}{n^{p+1}}with0<p<1.$$

Commented byabdo mathsup 649 cc last updated on 11/May/18

$$wehave\frac{n^p{sin}^2(n!)}{n^{p+1}}=\frac{{sin}^2(n!)}{n}but$$ $$0⩽{sin}^2(n!)⩽1\Rightarrow 0⩽\frac{{sin}^2(n!)}{n}⩽\frac{1}{n}\to 0(n\to +\mathrm{\infty })so$$ $${lim}_{n\to +\mathrm{\infty }}\frac{n^p{sin}^2(n!)}{n^{p+1}}=0.$$

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

$$=\frac{lim}{n\to +\mathrm{\infty }}\frac{{sin}^2(n!)}{n}$$ $$lett=\frac{1}{n}whenn\to +\mathrm{\infty }t\to +0$$ $$\frac{lim}{t\to +0}t.\{{sin}^2\left(\frac{1}{t}\right)!\}$$ $$=0\times anyvaluebetween+/-1$$ $$=0$$ $$=$$

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