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Question Number 98616 by bemath last updated on 15/Jun/20

Answered by bobhans last updated on 15/Jun/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\sqrt{5}}{\sqrt{9+9}}\right)}^{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\sqrt{5}}{\sqrt{18}}\right)}^{\mathrm{n}}$$$$=\sqrt{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{5}{18}\right)}^{\mathrm{n}}}=0.◼$$

Answered by mathmax by abdo last updated on 15/Jun/20

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}=\frac{{(2+\mathrm{i})}^{\mathrm{n}}}{{(3-3\mathrm{i})}^{\mathrm{n}}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{{\left(\sqrt{5}{\mathrm{e}}^{\mathrm{iarctan}\left(\frac{1}{2}\right)}\right)}^{\mathrm{n}}}{3^{\mathrm{n}}{\left(\sqrt{2}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}\right)}^{\mathrm{n}}}=\frac{{\left(\sqrt{5}\right)}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{inarctan}\left(\frac{1}{2}\right)}}{{\left(3\sqrt{2}\right)}^{\mathrm{n}}{\mathrm{e}}^{-\frac{\mathrm{in}\pi }{4}}}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}={\left(\frac{\sqrt{5}}{3\sqrt{2}}\right)}^{\mathrm{n}}{\mathrm{e}}^{\frac{\mathrm{in}\pi }{4}-\mathrm{inarctan}\left(\frac{1}{2}\right)}\Rightarrow \mid {\mathrm{A}}_{\mathrm{n}}\mid ={\left(\frac{\sqrt{5}}{3\sqrt{2}}\right)}^{\mathrm{n}}\to 0(\mathrm{n}\to \mathrm{\infty })\mathrm{due}\mathrm{to}\mid \frac{\sqrt{5}}{3\sqrt{2}}\mid <1$$$$\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=0$$

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