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Category: Integration

Question-60716

Question Number 60716 by peter frank last updated on 24/May/19 Commented by maxmathsup by imad last updated on 25/May/19 $${let}\:{A}\left({x}\right)\:=\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{{x}} \:\Rightarrow{A}\left({x}\right)\:=\left(\mathrm{1}−\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}} \:={e}^{{x}\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)} \\ $$$${but}\:{we}\:{have}\:{for}\:{u}\:\in{V}\left(\mathrm{0}\right)\:{ln}\left(\mathrm{1}−{u}\right)\:\sim−{u}\:\Rightarrow{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\sim−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{for}\:{x}\in{V}\left(+\infty\right) \\…