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Question Number 188343 by Shrinava last updated on 28/Feb/23
Find: Ω = Σ_(n=1) ^∞ (lim_(x→0) (1 - ((5 - (√(25 - x^2 )))/x) )^((5n)/x) )
$$\mathrm{Find}:\:\Omega\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}\:-\:\frac{\mathrm{5}\:-\:\sqrt{\mathrm{25}\:-\:\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}}\:\right)^{\frac{\mathrm{5}\boldsymbol{\mathrm{n}}}{\boldsymbol{\mathrm{x}}}} \:\right) \\ $$
Answered by SEKRET last updated on 28/Feb/23
lim_(x→0) (1−((5−(√(25−x^2 )))/x))^((5n)/x) =lim_(x→0) (1+(1/((x/( (√(25−x^2 )) −5 )) )))^((x/( (√(25−x^2 )) −5))∙((5n∙((√(25−x^2 )) −5))/x^2 )) (x/( (√(25−x^2 )) −5)) = a x→0 a→∓∞ lim_(a→∓∞) (1+(1/a))^(a ∙5n∙lim_(x→0) (((√(25−x^2 )) −5)/x^2 )) = e^((−5n)/(10)) Σ_(n=1) ^∞ e^((−n)/2) = ? e^((−1)/2) +e^((−2)/2) +e^((−3)/2) +e^((−4)/2) +e^((−5)/2) +.....=A e^((−1)/2) +e^((−1)/2) ∙(e^((−1)/2) +e^((−2)/2) +......)_(A) = A A= (e^((−1)/2) /(1−e^((−1)/2) ))= ((√e)/(e−(√e) )) = (1/( (√e) −1)) ABDULAZIZ ABDUVALIYEV
$$\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\left(\mathrm{1}−\frac{\mathrm{5}−\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}{\boldsymbol{\mathrm{x}}}\right)^{\frac{\mathrm{5}\boldsymbol{\mathrm{n}}}{\boldsymbol{\mathrm{x}}}} =\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\left(\mathrm{1}+\frac{\mathrm{1}}{\frac{\boldsymbol{\mathrm{x}}}{\:\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:−\mathrm{5}\:\:}\:\:}\right)^{\frac{\boldsymbol{\mathrm{x}}}{\:\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:−\mathrm{5}}\centerdot\frac{\mathrm{5}\boldsymbol{\mathrm{n}}\centerdot\left(\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\:−\mathrm{5}\right)}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }} \\ $$$$\:\:\frac{\boldsymbol{\mathrm{x}}}{\:\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\:−\mathrm{5}}\:=\:\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}\:\:\:\:\:\boldsymbol{\mathrm{a}}\rightarrow\mp\infty \\ $$$$\:\:\underset{\boldsymbol{\mathrm{a}}\rightarrow\mp\infty} {\boldsymbol{\mathrm{lim}}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\right)^{\boldsymbol{\mathrm{a}}\:\centerdot\mathrm{5}\boldsymbol{\mathrm{n}}\centerdot\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\frac{\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:−\mathrm{5}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }} =\:\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{5}\boldsymbol{\mathrm{n}}}{\mathrm{10}}} \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{e}}^{\frac{−\boldsymbol{\mathrm{n}}}{\mathrm{2}}} =\:? \\ $$$$\:\:\:\:\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{1}}{\mathrm{2}}} +\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{2}}{\mathrm{2}}} +\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{3}}{\mathrm{2}}} +\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{4}}{\mathrm{2}}} +\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{5}}{\mathrm{2}}} +…..=\boldsymbol{\mathrm{A}} \\ $$$$\:\:\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{1}}{\mathrm{2}}} +\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{1}}{\mathrm{2}}} \centerdot\underset{\boldsymbol{\mathrm{A}}} {\left(\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{1}}{\mathrm{2}}} +\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{2}}{\mathrm{2}}} +……\right)}=\:\boldsymbol{\mathrm{A}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}=\:\frac{\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−\boldsymbol{\mathrm{e}}^{\frac{−\mathrm{1}}{\mathrm{2}}} }=\:\frac{\sqrt{\boldsymbol{\mathrm{e}}}}{\boldsymbol{\mathrm{e}}−\sqrt{\boldsymbol{\mathrm{e}}}\:}\:=\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{e}}}\:−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{ABDULAZIZ}}\:\:\:\boldsymbol{\mathrm{ABDUVALIYEV}}\: \\ $$$$ \\ $$$$ \\ $$

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