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Question Number 54566 by mr W last updated on 06/Feb/19

$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{1+c}{1+{\mu }^{\frac{1}{n}}}\right)}^n=?$$ $$with0⩽c⩽1,\mu >0$$

Commented bymaxmathsup by imad last updated on 06/Feb/19

$$let{A}_n={\left(\frac{1+e}{1+{\mu }^{\frac{1}{n}}}\right)}^n\Rightarrow A_n={\left(\frac{1+e+{\mu }^{\frac{1}{n}}-{\mu }^{\frac{1}{n}}}{1+{\mu }^{\frac{1}{n}}}\right)}^n={(1+\frac{e-{\mu }^{\frac{1}{n}}}{1+{\mu }^{\frac{1}{n}}})}^n$$ $$=e^{nln(1+\frac{e-{\mu }^{\frac{1}{n}}}{1+{\mu }^{\frac{1}{n}}})but{\mu }^{\frac{1}{n}}=e^{\frac{ln\left(\mu \right)}{n}}\sim 1+\frac{ln\left(\mu \right)}{n}}\Rightarrow e-{\mu }^{\frac{1}{n}}\sim e-1-\frac{ln\left(\mu \right)}{n}and$$ $$1+{\mu }^{\frac{1}{n}}=2+\frac{ln\left(\mu \right)}{n}\Rightarrow 1+\frac{e-{\mu }^{\frac{1}{n}}}{1+{\mu }^{\frac{1}{n}}}\sim 1+\frac{e-1}{2}=\frac{1+e}{2}\Rightarrow A_n\sim e^{nln\left(\frac{1+e}{2}\right)}\to +\mathrm{\infty }(n\to +\mathrm{\infty })$$ $$(lookthatln\left(\frac{1+e}{2}\right)>0)weconsider\mu andefixed..$$ $$$$

Commented bymr W last updated on 06/Feb/19

$$thankyousir!$$ $$c⩽1$$ $$therefore\ln \left(\frac{1+c}{2}\right)⩽0$$ $$itseemsifc<1,{\left(\frac{1+e}{1+{\mu }^{\frac{1}{n}}}\right)}^n\to 0$$ $$whatifc=1?,i.e.{\left(\frac{2}{1+{\mu }^{\frac{1}{n}}}\right)}^n\to ?$$

Commented bymaxmathsup by imad last updated on 06/Feb/19

$$youarerightsirihavecommitedaerrorofcalculuswehave$$ $$0<e<1\Rightarrow 1<1+e<2\Rightarrow \frac{1}{2}<\frac{1+e}{2}<1\Rightarrow ln\left(\frac{1+e}{2}\right)<0\Rightarrow {lim}_{n\to +\mathrm{\infty }}{A}_n=0$$ $$forgivemeiwastired...$$

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