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

$$\mathrm{f}\left(0\right)=0$$$$\mathrm{f}\left(1\right)={\mathrm{e}}^3$$$$\mathrm{f}\in {\mathbb{C}}^2\left(\mathbb{R}\right)$$$$\mathrm{f}{}^{{}^{″}}\left(\mathrm{x}\right)-5\mathrm{f}{}^{{}^{\prime }}\left(\mathrm{x}\right)+6\mathrm{f}\left(\mathrm{x}\right)=0,\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$$$\mathrm{Find}:\mathbf{\Omega }=\underset{\mathbf{x}\to \mathrm{\infty }}{\lim }{(1-\frac{1}{\mathrm{f}\left(\mathrm{x}\right)})}^{\mathbf{x}}$$

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

$$r^2-5r+6=0$$$$(r-2)(r-3)=0$$$$r=2,3$$$$f\left(x\right)=C_1{e}^{2x}+C_2{e}^{3x}$$$$f\left(0\right)=C_1+C_2=0\Rightarrow C_1=-C_2$$$$f\left(1\right)=C_1{e}^2+C_2{e}^3=e^3\Rightarrow C_1+C_{2}e=e$$$$\Rightarrow C_2=\frac{e}{e-1}=-C_1$$$$\Rightarrow f\left(x\right)=\frac{e}{e-1}(-e^{2x}+e^{3x})$$$$\mathrm{\Omega }=\underset{x\to \mathrm{\infty }}{\lim }{(1-\frac{1}{f\left(x\right)})}^x$$$$=\underset{x\to \mathrm{\infty }}{\lim }{(1-\frac{e-1}{e^{2x+1}(e^x-1)})}^x=1$$

Commented by Shrinava last updated on 01/Dec/22

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

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