Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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 )))=−∞

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com