Question Number 218185 by hardmath last updated on 31/Mar/25
$$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:\left(\mathrm{n}\pi\:\sqrt{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{2n}\:+\:\mathrm{2}\centerdot\left(\mathrm{k}\:+\:\mathrm{1}\right)}\right)\:=\:? \\ $$$$\mathrm{k}\:\in\:\mathbb{Z}\:-\:\mathrm{fixed} \\ $$
Answered by vnm last updated on 01/Apr/25
$${n}\pi\sqrt{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{2}\left({k}+\mathrm{1}\right)}=\sqrt{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}+\mathrm{2}{k}+\mathrm{1}}= \\ $$$${n}\pi\sqrt{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}{k}+\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right)}= \\ $$$${n}\pi\left({n}+\mathrm{1}\right)\left(\mathrm{1}+\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\right)= \\ $$$$\pi\left({n}\left({n}+\mathrm{1}\right)+\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)+{o}\left(\mathrm{1}\right)\right)= \\ $$$$\pi\left(\mathrm{2}{m}+\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)+{o}\left(\mathrm{1}\right)\right)={A},\:{m}\in\mathbb{N} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{A}=\mathrm{sin}\:\left(\pi{k}+\frac{\pi}{\mathrm{2}}\right)=\left(−\mathrm{1}\right)^{{k}} \\ $$
Commented by hardmath last updated on 01/Apr/25
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor} \\ $$$$\mathrm{answer}\:\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} \:=\:? \\ $$
Commented by vnm last updated on 01/Apr/25
$$\left(−\mathrm{1}\right)^{{k}} \:{is}\:{the}\:{answer}. \\ $$$$\left(−\mathrm{1}\right)^{{k}} =\mathrm{1}\:{if}\:{k}\:{is}\:{even}\:{and}\:=−\mathrm{1}\:{if}\:{odd}. \\ $$
Commented by hardmath last updated on 03/Apr/25
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$