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Question Number 79679 by ~blr237~ last updated on 27/Jan/20

Find lim_(x→∞)  x^2 ln(1+x)−x

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

Commented by mathmax by abdo last updated on 27/Jan/20

let f(x)=x^2 ln(1+x)−x ⇒f(x)=x^2 ln(x(1+(1/x)))−x  =x^2 lnx +x^2 ln(1+(1/x))−x∼x^2 lnx+x^2 ((1/x))−x ⇒  f(x)∼x^2 lnx ⇒lim_(x→+∞) f(x)=+∞

$${let}\:{f}\left({x}\right)={x}^{\mathrm{2}} {ln}\left(\mathrm{1}+{x}\right)−{x}\:\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}} {ln}\left({x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right)−{x} \\ $$$$={x}^{\mathrm{2}} {lnx}\:+{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\sim{x}^{\mathrm{2}} {lnx}+{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)−{x}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim{x}^{\mathrm{2}} {lnx}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty \\ $$$$ \\ $$

Commented by ~blr237~ last updated on 27/Jan/20

thanks sir

$${thanks}\:{sir} \\ $$

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