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Question Number 196321 by sniper237 last updated on 22/Aug/23

$$\underset{n\to +\mathrm{\infty }}{\lim }sin\left(2\pi \sqrt{n^2+1}\right)=0$$$$\underset{n\to +\mathrm{\infty }}{\lim }arg(n^2+n+1+i)=0$$

Answered by witcher3 last updated on 22/Aug/23

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim sin}\left(2\pi \sqrt{{\mathrm{n}}^2+1}\right)..$$$$\sin \left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)=\sin \left(2\pi \mathrm{n}\left(\sqrt{1+\frac{1}{{\mathrm{n}}^2}}\right)\right)$$$$\sqrt{1+\frac{1}{{\mathrm{n}}^2}}=1+\frac{1}{{2\mathrm{n}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)$$$$\sin \left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)=\sin (2\pi +\frac{\pi }{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right))$$$$=\sin (\frac{\pi }{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)\to 0$$$$\mathrm{elementry}\mathrm{why}$$$$\sin (2\pi \mathrm{n}+\mathrm{x})=\sin \left(\mathrm{x}\right),\mathrm{\forall }(\mathrm{n},\mathrm{x})\in \mathbb{Z}\ast \mathbb{C}$$$$\iff \underset{\mathrm{n}\to \mathrm{\infty }}{\lim sin}\left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim sin}(2\pi \sqrt{1+{\mathrm{n}}^2}-2\pi \mathrm{n})$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim sin}\left(\frac{2\pi }{\sqrt{1+{\mathrm{n}}^2}+\mathrm{n}}\right)=\sin (\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }.\frac{2\pi }{\mathrm{n}+\sqrt{1+{\mathrm{n}}^2}})$$$$\mathrm{by}\mathrm{continuity}\mathrm{of}\sin$$$$=\sin \left(0\right)=0$$$$\arg ({\mathrm{n}}^2+\mathrm{n}+1+\mathrm{i})\equiv {\tan }^{-1}\left(\frac{1}{1+\mathrm{n}+{\mathrm{n}}^2}\right)\left[2\pi \right]$$$${\underset{\mathrm{n}\to \mathrm{\infty }}{\lim tan}}^{-1}\left(\frac{1}{1+\mathrm{n}+{\mathrm{n}}^2}\right)={\tan }^{-1}\left(\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{{\mathrm{n}}^2+\mathrm{n}+1}\right)]$$$$\equiv 0\left[2\pi \right]$$$$$$

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