Question Number 196267 by Rodier97 last updated on 21/Aug/23
$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:{if}\:\:\:{lim}_{{n}\rightarrow+\infty} \:\left(\mathrm{1}+\:\frac{{x}}{\mathrm{7}{n}}\right)^{\mathrm{29}{n}} =\mathrm{2023} \\ $$$$\:\:\:\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}}\:?? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 21/Aug/23
$${find}\:{x}? \\ $$$${but}\:{you}\:{said}\:{x}\rightarrow+\infty \\ $$
Commented by mr W last updated on 21/Aug/23
$${i}\:{guess}\:{you}\:{meant}\:{n}\rightarrow+\infty \\ $$
Answered by mr W last updated on 21/Aug/23
![lim_(n→+∞) (1+ (x/(7n)))^(29n) =lim_(n→+∞) [(1+ (x/(7n)))^((7n)/x) ]^((29x)/7) =lim_(m→+∞) [(1+ (1/m))^m ]^((29x)/7) with m=((7n)/x) =e^((29x)/7) =2023 ⇒x=(7/(29))ln (2023)](Q196270.png)
$$\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\:\frac{{x}}{\mathrm{7}{n}}\right)^{\mathrm{29}{n}} \\ $$$$\:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\left[\left(\mathrm{1}+\:\frac{{x}}{\mathrm{7}{n}}\right)^{\frac{\mathrm{7}{n}}{{x}}} \right]^{\frac{\mathrm{29}{x}}{\mathrm{7}}} \\ $$$$\:=\underset{{m}\rightarrow+\infty} {\mathrm{lim}}\:\left[\left(\mathrm{1}+\:\frac{\mathrm{1}}{{m}}\right)^{{m}} \right]^{\frac{\mathrm{29}{x}}{\mathrm{7}}} \:\:{with}\:{m}=\frac{\mathrm{7}{n}}{{x}} \\ $$$$={e}^{\frac{\mathrm{29}{x}}{\mathrm{7}}} =\mathrm{2023} \\ $$$$\Rightarrow{x}=\frac{\mathrm{7}}{\mathrm{29}}\mathrm{ln}\:\left(\mathrm{2023}\right) \\ $$