Question Number 221377 by hardmath last updated on 01/Jun/25
$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{10}}\:−\:\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \:=\:? \\ $$
Answered by Ghisom last updated on 01/Jun/25
![ln Ω =lim_(n→∞) nln (2×10^(1/n) −1) n=(1/k) ln Ω =lim_(k→0^+ ) ((ln (2×10^k −1))/k) [l′Ho^ pital] ln Ω =lim_(k→0^+ ) ((2×10^k ln 10)/(2×10^k −1)) =2ln 10 Ω=100](https://www.tinkutara.com/question/Q221378.png)
$$\mathrm{ln}\:\Omega\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\mathrm{ln}\:\left(\mathrm{2}×\mathrm{10}^{\mathrm{1}/{n}} −\mathrm{1}\right) \\ $$$${n}=\frac{\mathrm{1}}{{k}} \\ $$$$\mathrm{ln}\:\Omega\:=\underset{{k}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{2}×\mathrm{10}^{{k}} −\mathrm{1}\right)}{{k}} \\ $$$$\:\:\:\:\:\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$\mathrm{ln}\:\Omega\:=\underset{{k}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{2}×\mathrm{10}^{{k}} \mathrm{ln}\:\mathrm{10}}{\mathrm{2}×\mathrm{10}^{{k}} −\mathrm{1}}\:=\mathrm{2ln}\:\mathrm{10} \\ $$$$\Omega=\mathrm{100} \\ $$
Answered by vnm last updated on 02/Jun/25
![2((10))^(1/n) −1=2e^((ln 10)/n) −1=2(1+((ln 10)/n)+o((1/n)))−1= 1+((2ln 10)/n)+o((1/n))=1+((2ln 10)/n)+((α(n))/n), α(n)→_(n→∞) 0 (1+((2ln 10+α(n))/n))^n =[(1+((2ln 10+α(n))/n))^(n/(2ln 10+α(n))) ]^(2ln 10+α(n)) →_(n→∞) e^(2ln 10+0) =100](https://www.tinkutara.com/question/Q221381.png)
$$\mathrm{2}\sqrt[{{n}}]{\mathrm{10}}−\mathrm{1}=\mathrm{2}{e}^{\frac{\mathrm{ln}\:\mathrm{10}}{{n}}} −\mathrm{1}=\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{ln}\:\mathrm{10}}{{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)−\mathrm{1}= \\ $$$$\mathrm{1}+\frac{\mathrm{2ln}\:\mathrm{10}}{{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)=\mathrm{1}+\frac{\mathrm{2ln}\:\mathrm{10}}{{n}}+\frac{\alpha\left({n}\right)}{{n}},\:\alpha\left({n}\right)\underset{{n}\rightarrow\infty} {\rightarrow}\mathrm{0} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{2ln}\:\mathrm{10}+\alpha\left({n}\right)}{{n}}\right)^{{n}} =\left[\left(\mathrm{1}+\frac{\mathrm{2ln}\:\mathrm{10}+\alpha\left({n}\right)}{{n}}\right)^{\frac{{n}}{\mathrm{2ln}\:\mathrm{10}+\alpha\left({n}\right)}} \right]^{\mathrm{2ln}\:\mathrm{10}+\alpha\left({n}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\rightarrow}{e}^{\mathrm{2ln}\:\mathrm{10}+\mathrm{0}} =\mathrm{100} \\ $$