Question Number 33073 by NECx last updated on 10/Apr/18
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {e}^{−{x}} \\ $$
Commented by abdo imad last updated on 10/Apr/18
$${for}\:{all}\:{plynome}\:{p}\left({x}\right)\:{not}\:{o}\:{wehave}\:{lim}_{{x}\rightarrow+\infty} {p}\left({x}\right){e}^{−\alpha{x}} =\mathrm{0} \\ $$$${with}\:\alpha>\mathrm{0}\:\:{we}\:{say}\:{that}\:{e}^{−\alpha{x}} \:{defeat}\:{p}\left({x}\right)\:{at}\:+\infty\:. \\ $$
Answered by MJS last updated on 10/Apr/18
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{{p}} }{{q}^{{x}} }=\mathrm{0}\:\forall{q}>\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{\mathrm{log}_{\mathrm{10}} \:\left(\frac{{x}^{{p}} }{{q}^{{x}} }\right)}{{x}}=\frac{{p}\mathrm{log}_{\mathrm{10}} \:{x}}{{x}}−\mathrm{log}_{\mathrm{10}} \:{q} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{f}\left({x}\right)=−\mathrm{log}_{\mathrm{10}} \:{q}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{xf}\left({x}\right)=−\infty\:\Rightarrow \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{10}^{{xf}\left({x}\right)} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{{p}} }{{q}^{{x}} }=\mathrm{0} \\ $$$$ \\ $$
Answered by Joel578 last updated on 10/Apr/18
$${L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{{e}^{{x}} } \\ $$$$\:\:\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}{x}}{{e}^{{x}} }\:\:\:\left({L}'{Hospital}\right) \\ $$$$\:\:\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}}{{e}^{{x}} } \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{2}}{\infty}\:=\:\mathrm{0} \\ $$