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Question Number 33073 by NECx last updated on 10/Apr/18

$$\underset{x\to \mathrm{\infty }}{\lim }{x}^2{e}^{-x}$$

Commented by abdo imad last updated on 10/Apr/18

$$forallplynomep\left(x\right)notowehave{lim}_{x\to +\mathrm{\infty }}p\left(x\right)e^{-\alpha x}=0$$$$with\alpha >0wesaythat{e}^{-\alpha x}defeatp\left(x\right)at+\mathrm{\infty }.$$

Answered by MJS last updated on 10/Apr/18

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{x^p}{q^x}=0\mathrm{\forall }q>1$$$$f\left(x\right)=\frac{{\log }_{10}\left(\frac{x^p}{q^x}\right)}{x}=\frac{p{\log }_{10}x}{x}-{\log }_{10}q$$$$\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=-{\log }_{10}q\Rightarrow$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }xf\left(x\right)=-\mathrm{\infty }\Rightarrow$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }{10}^{xf\left(x\right)}=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^p}{q^x}=0$$$$$$

Answered by Joel578 last updated on 10/Apr/18

$$L=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^2}{e^x}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{2x}{e^x}\left(L^{\prime }Hospital\right)$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{e^x}$$$$=\frac{2}{\mathrm{\infty }}=0$$

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