Question Number 181794 by Shrinava last updated on 30/Nov/22
$$\mathrm{f}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)\:=\:\mathrm{e}^{\mathrm{3}} \\ $$$$\mathrm{f}\:\in\:\mathbb{C}^{\mathrm{2}} \:\left(\mathbb{R}\right) \\ $$$$\mathrm{f}\:^{''} \:\left(\mathrm{x}\right)\:−\:\mathrm{5}\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\:+\:\mathrm{6}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{0}\:\:,\:\:\forall\:\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{f}\:\left(\mathrm{x}\right)}\right)^{\boldsymbol{\mathrm{x}}} \\ $$
Answered by mr W last updated on 01/Dec/22
$${r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{6}=\mathrm{0} \\ $$$$\left({r}−\mathrm{2}\right)\left({r}−\mathrm{3}\right)=\mathrm{0} \\ $$$${r}=\mathrm{2},\:\mathrm{3} \\ $$$${f}\left({x}\right)={C}_{\mathrm{1}} {e}^{\mathrm{2}{x}} +{C}_{\mathrm{2}} {e}^{\mathrm{3}{x}} \\ $$$${f}\left(\mathrm{0}\right)={C}_{\mathrm{1}} +{C}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow{C}_{\mathrm{1}} =−{C}_{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)={C}_{\mathrm{1}} {e}^{\mathrm{2}} +{C}_{\mathrm{2}} {e}^{\mathrm{3}} ={e}^{\mathrm{3}} \:\Rightarrow{C}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}={e} \\ $$$$\:\Rightarrow{C}_{\mathrm{2}} =\frac{{e}}{{e}−\mathrm{1}}=−{C}_{\mathrm{1}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{{e}}{{e}−\mathrm{1}}\left(−{e}^{\mathrm{2}{x}} +{e}^{\mathrm{3}{x}} \right) \\ $$$$\Omega=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{{x}} \\ $$$$\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{{e}−\mathrm{1}}{{e}^{\mathrm{2}{x}+\mathrm{1}} \left({e}^{{x}} −\mathrm{1}\right)}\right)^{{x}} =\mathrm{1} \\ $$
Commented by Shrinava last updated on 01/Dec/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$