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

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

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

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 ��

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