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Question Number 79689 by ~blr237~ last updated on 27/Jan/20

Find lim_(x→∞) x^2 ln(1+(1/x)) −x without DL and if you have to use the hospital rule please justify that your function is C^1

$${Find}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:−{x}\:\:\:\:\:{without}\:\:{DL}\:{and}\:{if}\:{you}\:{have}\:{to}\:{use}\:{the}\: \\ $$$${hospital}\:{rule}\:{please}\:\:{justify}\:{that}\:{your}\:{function}\:{is}\:{C}^{\mathrm{1}} \: \\ $$

Commented by mr W last updated on 27/Jan/20

lim_(x→∞) x^2 ln(1+(1/x)) −x =lim_(x→0) (1/x^2 )ln(1+x) −(1/x) =lim_(x→0) (1/x^2 )(x−(x^2 /2)+(x^3 /3)−(x^4 /4)+...) −(1/x) =lim_(x→0) (−(1/2)+(x/3)−(x^2 /4)+...) =−(1/2)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:−{x} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{x}\right)\:−\frac{\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left({x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+...\right)\:−\frac{\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{{x}}{\mathrm{3}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+...\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by john santu last updated on 27/Jan/20

mac′laurin series

$$\mathrm{mac}'\mathrm{laurin}\:\mathrm{series} \\ $$

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