Question Number 34774 by abdo mathsup 649 cc last updated on 11/May/18
$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\frac{{n}^{{p}} \:{sin}^{\mathrm{2}} \left({n}!\right)}{{n}^{{p}+\mathrm{1}} }\:\:{with}\mathrm{0}<{p}<\mathrm{1}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 11/May/18
$${we}\:{have}\:\frac{{n}^{{p}} \:{sin}^{\mathrm{2}} \left({n}!\right)}{{n}^{{p}+\mathrm{1}} }\:=\:\frac{{sin}^{\mathrm{2}} \left({n}!\right)}{{n}}\:{but} \\ $$$$\mathrm{0}\leqslant{sin}^{\mathrm{2}} \left({n}!\right)\leqslant\mathrm{1}\:\Rightarrow\:\mathrm{0}\leqslant\:\frac{{sin}^{\mathrm{2}} \left({n}!\right)}{{n}}\leqslant\frac{\mathrm{1}}{{n}}\:\rightarrow\mathrm{0}\left({n}\rightarrow+\infty\right){so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\frac{{n}^{{p}} \:{sin}^{\mathrm{2}} \left({n}!\right)}{{n}^{{p}+\mathrm{1}} }\:=\mathrm{0}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/May/18
$$=\frac{{lim}}{{n}\rightarrow+\infty}\frac{{sin}^{\mathrm{2}} \left({n}!\right)}{{n}} \\ $$$${let}\:{t}=\frac{\mathrm{1}}{{n}}{when}\:{n}\rightarrow+\infty\:\:{t}\rightarrow+\mathrm{0} \\ $$$$\frac{{lim}}{{t}\rightarrow+\mathrm{0}}\:{t}.\left\{{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{t}}\right)!\right\} \\ $$$$=\mathrm{0}×{any}\:{value}\:{between}\:+/−\:\:\mathrm{1} \\ $$$$=\mathrm{0} \\ $$$$= \\ $$