Question Number 196321 by sniper237 last updated on 22/Aug/23
$$\:\:\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{sin}\left(\mathrm{2}\pi\sqrt{{n}^{\mathrm{2}} +\mathrm{1}\:}\:\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\:{arg}\left({n}^{\mathrm{2}} +{n}+\mathrm{1}+{i}\right)\:=\:\mathrm{0} \\ $$
Answered by witcher3 last updated on 22/Aug/23
![lim_(n→∞) sin(2π(√(n^2 +1))).. sin(2π(√(1+n^2 )))=sin(2πn((√(1+(1/n^2 ))))) (√(1+(1/n^2 )))=1+(1/(2n^2 ))+o((1/n^2 )) sin(2π(√(1+n^2 )))=sin(2π+(π/n)+o((1/n))) =sin((π/n)+o((1/n))→0 elementry why sin(2πn+x)=sin(x),∀(n,x)∈Z∗C ⇔lim_(n→∞) sin(2π(√(1+n^2 )))=lim_(n→∞) sin(2π(√(1+n^2 ))−2πn) =lim_(n→∞) sin(((2π)/( (√(1+n^2 ))+n)))=sin(lim_(n→∞) .((2π)/(n+(√(1+n^2 ))))) by continuity of sin =sin(0)=0 arg(n^2 +n+1+i)≡tan^(−1) ((1/(1+n+n^2 )))[2π] lim_(n→∞) tan^(−1) ((1/(1+n+n^2 )))=tan^(−1) (lim_(n→∞) (1/(n^2 +n+1)))] ≡0[2π]](https://www.tinkutara.com/question/Q196330.png)
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}sin}\left(\mathrm{2}\pi\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}\right).. \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)=\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }}\right)\right) \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)=\mathrm{sin}\left(\mathrm{2}\pi+\frac{\pi}{\mathrm{n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right) \\ $$$$=\mathrm{sin}\left(\frac{\pi}{\mathrm{n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\rightarrow\mathrm{0}\right. \\ $$$$\mathrm{elementry}\:\mathrm{why} \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}+\mathrm{x}\right)=\mathrm{sin}\left(\mathrm{x}\right),\forall\left(\mathrm{n},\mathrm{x}\right)\in\mathbb{Z}\ast\mathbb{C} \\ $$$$\Leftrightarrow\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }−\mathrm{2}\pi\mathrm{n}\right) \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}sin}\left(\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }+\mathrm{n}}\right)=\mathrm{sin}\left(\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{2}\pi}{\mathrm{n}+\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{by}\:\mathrm{continuity}\:\mathrm{of}\:\mathrm{sin} \\ $$$$=\mathrm{sin}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{arg}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}+\mathrm{i}\right)\equiv\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{n}+\mathrm{n}^{\mathrm{2}} }\right)\left[\mathrm{2}\pi\right] \\ $$$$\left.\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{n}+\mathrm{n}^{\mathrm{2}} }\right)=\mathrm{tan}^{−\mathrm{1}} \left(\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}}\right)\right] \\ $$$$\equiv\mathrm{0}\left[\mathrm{2}\pi\right] \\ $$$$ \\ $$