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Question Number 181794 by Shrinava last updated on 30/Nov/22

f(0) = 0 f(1) = e^3 f ∈ C^2 (R) f^(′′) (x) − 5 f^′ (x) + 6 f(x) = 0 , ∀ x ∈ R Find: 𝛀 =lim_(x→∞) (1 − (1/(f (x))))^x

$$\mathrm{f}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)\:=\:\mathrm{e}^{\mathrm{3}} \\ $$$$\mathrm{f}\:\in\:\mathbb{C}^{\mathrm{2}} \:\left(\mathbb{R}\right) \\ $$$$\mathrm{f}\:^{''} \:\left(\mathrm{x}\right)\:−\:\mathrm{5}\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\:+\:\mathrm{6}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{0}\:\:,\:\:\forall\:\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{f}\:\left(\mathrm{x}\right)}\right)^{\boldsymbol{\mathrm{x}}} \\ $$

Answered by mr W last updated on 01/Dec/22

r^2 −5r+6=0 (r−2)(r−3)=0 r=2, 3 f(x)=C_1 e^(2x) +C_2 e^(3x) f(0)=C_1 +C_2 =0 ⇒C_1 =−C_2 f(1)=C_1 e^2 +C_2 e^3 =e^3 ⇒C_1 +C_2 e=e ⇒C_2 =(e/(e−1))=−C_1 ⇒f(x)=(e/(e−1))(−e^(2x) +e^(3x) ) Ω=lim_(x→∞) (1−(1/(f(x))))^x =lim_(x→∞) (1−((e−1)/(e^(2x+1) (e^x −1))))^x =1

$${r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{6}=\mathrm{0} \\ $$$$\left({r}−\mathrm{2}\right)\left({r}−\mathrm{3}\right)=\mathrm{0} \\ $$$${r}=\mathrm{2},\:\mathrm{3} \\ $$$${f}\left({x}\right)={C}_{\mathrm{1}} {e}^{\mathrm{2}{x}} +{C}_{\mathrm{2}} {e}^{\mathrm{3}{x}} \\ $$$${f}\left(\mathrm{0}\right)={C}_{\mathrm{1}} +{C}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow{C}_{\mathrm{1}} =−{C}_{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)={C}_{\mathrm{1}} {e}^{\mathrm{2}} +{C}_{\mathrm{2}} {e}^{\mathrm{3}} ={e}^{\mathrm{3}} \:\Rightarrow{C}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}={e} \\ $$$$\:\Rightarrow{C}_{\mathrm{2}} =\frac{{e}}{{e}−\mathrm{1}}=−{C}_{\mathrm{1}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{{e}}{{e}−\mathrm{1}}\left(−{e}^{\mathrm{2}{x}} +{e}^{\mathrm{3}{x}} \right) \\ $$$$\Omega=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{{x}} \\ $$$$\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{{e}−\mathrm{1}}{{e}^{\mathrm{2}{x}+\mathrm{1}} \left({e}^{{x}} −\mathrm{1}\right)}\right)^{{x}} =\mathrm{1} \\ $$

Commented by Shrinava last updated on 01/Dec/22

thank you dear professor cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$

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