lim-x-x-x-2-ln-1-1-x-

Question Number 72398 by 20190927 last updated on 28/Oct/19

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{x}−\mathrm{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\right] \\ $$

Commented by mathmax by abdo last updated on 28/Oct/19

let f(x)=x−x^2 ln(1+(1/x)) we have ln^′ (1+u)=(1/(1+u)) =1−u +u^2 +o(u^3 ) ⇒ ln(1+u)=u−(u^2 /2) +(u^3 /3) +o(u^4 ) (u→0) ⇒ ln(1+(1/x))=(1/x)−(1/(2x^2 )) +(1/(3x^3 )) +o((1/x^4 )) (x→+∞) ⇒ x^2 ln(1+(1/x))=x−(1/2) +(1/(3x)) +o((1/x^2 )) ⇒x−x^2 ln(1+(1/x)) =x−x+(1/2)−(1/(3x)) +o((1/x^2 )) ⇒f(x)∼(1/2)−(1/(3x)) (x→+∞) ⇒lim_(x→+∞) f(x)=(1/2)

$${let}\:{f}\left({x}\right)={x}−{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{u}^{\mathrm{2}} \:+{o}\left({u}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\:+{o}\left({u}^{\mathrm{4}} \right)\:\:\:\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)={x}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow{x}−{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$={x}−{x}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow{f}\left({x}\right)\sim\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{x}}\:\:\:\left({x}\rightarrow+\infty\right)\: \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 28/Oct/19