Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 92167 by john santu last updated on 05/May/20

lim_(x→∞) x−x^2 ln (1+(1/x)) ?

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

Commented by mathmax by abdo last updated on 06/May/20

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

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

Answered by jagoll last updated on 05/May/20

ln(1+(1/x)) = (1/x)−(1/(2x^2 ))+(1/(3x^3 ))−(1/(4x^4 ))+... ∣(1/x)∣ < 1 ⇒lim_(x→∞) x−x^2 ((1/x)−(1/(2x^2 ))+(1/(3x^3 ))−(1/(4x^4 ))+...)= lim_(x→∞) x−(x−(1/2)+(1/(3x))−(1/(4x^2 ))+...) = lim_(x→∞) (1/2)−(1/(3x))+(1/(4x^2 ))−... = (1/2)

$$\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{4}} }+... \\ $$$$\mid\frac{\mathrm{1}}{\mathrm{x}}\mid\:<\:\mathrm{1}\: \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}x}−\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{4}} }+...\right)= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}−\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3x}}−\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }+...\right)\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3x}}+\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }−...\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by john santu last updated on 05/May/20

cool man ��������

Terms of Service

Privacy Policy

Contact: info@tinkutara.com