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Question Number 167739 by DAVONG last updated on 24/Mar/22

Commented by mkam last updated on 24/Mar/22

−∞

$$−\infty \\ $$

Commented by mkam last updated on 24/Mar/22

−∞

$$−\infty \\ $$

Answered by Nimatullah last updated on 24/Mar/22

1

$$\mathrm{1} \\ $$

Answered by LEKOUMA last updated on 24/Mar/22

(x−2)e^x =−x−(x^3 /(3!))−2+o(x^3 )=−x−(x^3 /6)−2+o(x^3 ) (e^x −1)^3 =x^2 +x^3 +o(x^3 ) lim_(x→0) ((−2x−(x^3 /6))/(x^2 +x^3 ))=lim_(x→0) ((−2−(x^2 /6))/(x+x^2 ))=((−2)/0^+ )=−∞ Proposition 1

$$\left({x}−\mathrm{2}\right){e}^{{x}} =−{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−\mathrm{2}+{o}\left({x}^{\mathrm{3}} \right)=−{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−\mathrm{2}+{o}\left({x}^{\mathrm{3}} \right) \\ $$$$\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{3}} ={x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{3}} \right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2}{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}{{x}+{x}^{\mathrm{2}} }=\frac{−\mathrm{2}}{\mathrm{0}^{+} }=−\infty \\ $$$${Proposition}\:\mathrm{1} \\ $$

Answered by LEKOUMA last updated on 24/Mar/22

lim_(x→0) ((((x−2)e^x −(x−2))/((e^x −1)^3 )))=lim_(x→0) ((((x−2)(e^x −1))/((e^x −1)^3 ))) lim_(x→0) ((x−2)/((e^x −1)^2 ))=lim_(x→0) ((x−2)/((x+(x^2 /(2!)))^2 )) lim_(x→0) ((x−2)/x^2 )=((−2)/0^+ )−∞ Proposition 2

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\left({x}−\mathrm{2}\right){e}^{{x}} −\left({x}−\mathrm{2}\right)}{\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{3}} }\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\left({x}−\mathrm{2}\right)\left({e}^{{x}} −\mathrm{1}\right)}{\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{3}} }\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\mathrm{2}}{\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\mathrm{2}}{\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}\right)^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} }=\frac{−\mathrm{2}}{\mathrm{0}^{+} }−\infty \\ $$$${Proposition}\:\mathrm{2} \\ $$

Answered by greogoury55 last updated on 24/Mar/22

by L′Hopital lim_(x→0) (((e^x +(x−2)e^x −1)/(3(e^x −1)^2 ))) = −(2/3)×lim_(x→0) ((1/((e^x −1)^2 )))=−∞

$${by}\:{L}'{Hopital} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{e}^{{x}} +\left({x}−\mathrm{2}\right){e}^{{x}} −\mathrm{1}}{\mathrm{3}\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\:−\frac{\mathrm{2}}{\mathrm{3}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} }\right)=−\infty \\ $$

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