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Question Number 188343 by Shrinava last updated on 28/Feb/23

$$\mathrm{Find}:\mathrm{\Omega }=\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}\left(\underset{\mathbf{x}\to 0}{\lim }{(1-\frac{5-\sqrt{25-{\mathrm{x}}^2}}{\mathrm{x}})}^{\frac{5\mathbf{n}}{\mathbf{x}}}\right)$$

Answered by SEKRET last updated on 28/Feb/23

$$\underset{\mathbf{x}\to 0}{\mathbf{lim}}{(1-\frac{5-\sqrt{25-{\mathbf{x}}^2}}{\mathbf{x}})}^{\frac{5\mathbf{n}}{\mathbf{x}}}=\underset{\mathbf{x}\to 0}{\mathbf{lim}}{(1+\frac{1}{\frac{\mathbf{x}}{\sqrt{25-{\mathbf{x}}^2}-5}})}^{\frac{\mathbf{x}}{\sqrt{25-{\mathbf{x}}^2}-5}\cdot \frac{5\mathbf{n}\cdot (\sqrt{25-{\mathbf{x}}^2}-5)}{{\mathbf{x}}^2}}$$$$\frac{\mathbf{x}}{\sqrt{25-{\mathbf{x}}^2}-5}=\mathbf{a}\mathbf{x}\to 0\mathbf{a}\to \mp \mathrm{\infty }$$$$\underset{\mathbf{a}\to \mp \mathrm{\infty }}{\mathbf{lim}}{(1+\frac{1}{\mathbf{a}})}^{\mathbf{a}\cdot 5\mathbf{n}\cdot \underset{\mathbf{x}\to 0}{\mathbf{lim}}\frac{\sqrt{25-{\mathbf{x}}^2}-5}{{\mathbf{x}}^2}}={\mathbf{e}}^{\frac{-5\mathbf{n}}{10}}$$$$\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}{\mathbf{e}}^{\frac{-\mathbf{n}}{2}}=?$$$${\mathbf{e}}^{\frac{-1}{2}}+{\mathbf{e}}^{\frac{-2}{2}}+{\mathbf{e}}^{\frac{-3}{2}}+{\mathbf{e}}^{\frac{-4}{2}}+{\mathbf{e}}^{\frac{-5}{2}}+.....=\mathbf{A}$$$${\mathbf{e}}^{\frac{-1}{2}}+{\mathbf{e}}^{\frac{-1}{2}}\cdot \underset{\mathbf{A}}{({\mathbf{e}}^{\frac{-1}{2}}+{\mathbf{e}}^{\frac{-2}{2}}+......)}=\mathbf{A}$$$$\mathbf{A}=\frac{{\mathbf{e}}^{\frac{-1}{2}}}{1-{\mathbf{e}}^{\frac{-1}{2}}}=\frac{\sqrt{\mathbf{e}}}{\mathbf{e}-\sqrt{\mathbf{e}}}=\frac{1}{\sqrt{\mathbf{e}}-1}$$$$\mathbf{ABDULAZIZ}\mathbf{ABDUVALIYEV}$$$$$$$$$$

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