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 32991 by abdo imad last updated on 09/Apr/18

calculate lim_(x→0) ((ln(1+x) −x)/x^2 ) .

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)\:−{x}}{{x}^{\mathrm{2}} }\:. \\ $$

Commented by prof Abdo imad last updated on 09/Apr/18

for ∣x∣<1 (ln(1+x))^′ = (1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ ln(1+x)= Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n = x −(x^2 /2) +o(x^3 )⇒ ln(1+x)−x = −(x^2 /2)+o(x^3 ) ⇒ ((ln(1+x)−x)/x^2 ) =−(1/2) +o(x)⇒ lim_(x→0) ((ln(1+x)−x)/x^2 ) =((−1)/2) .

$${for}\:\mid{x}\mid<\mathrm{1}\:\:\:\left({ln}\left(\mathrm{1}+{x}\right)\right)^{'} =\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}} \\ $$$$\Rightarrow\:{ln}\left(\mathrm{1}+{x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \\ $$$$=\:{x}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:+{o}\left({x}^{\mathrm{3}} \right)\Rightarrow\:{ln}\left(\mathrm{1}+{x}\right)−{x}\:=\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\:\frac{{ln}\left(\mathrm{1}+{x}\right)−{x}}{{x}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\:+{o}\left({x}\right)\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{ln}\left(\mathrm{1}+{x}\right)−{x}}{{x}^{\mathrm{2}} }\:=\frac{−\mathrm{1}}{\mathrm{2}}\:. \\ $$

Answered by kyle_TW last updated on 09/Apr/18

L′Ho^(�) pital′s rule lim_(x→0) ((ln(1+x)−x)/x^2 ) =lim_(x→0) (((ln(1+x)−x)′)/((x^2 )′)) =lim_(x→0) ((((1/(1+x))−1)/(2x))) =lim_(x→0) (((1−1+x)/(2x(1+x)))) =lim_(x→0) ((1/(2(1+x))))= (1/2)

$${L}'{H}\overset{} {{o}pital}'{s}\:{rule} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left(\mathrm{1}+{x}\right)−{x}}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left({ln}\left(\mathrm{1}+{x}\right)−{x}\right)'}{\left({x}^{\mathrm{2}} \right)'} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}}−\mathrm{1}}{\mathrm{2}{x}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}−\mathrm{1}+{x}}{\mathrm{2}{x}\left(\mathrm{1}+{x}\right)}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{x}\right)}\right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by MJS last updated on 09/Apr/18

you made one small mistake (((1/(1+x))−1)/(2x))=((1−1−x)/(2x(1+x)))

$$\mathrm{you}\:\mathrm{made}\:\mathrm{one}\:\mathrm{small}\:\mathrm{mistake} \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}}−\mathrm{1}}{\mathrm{2}{x}}=\frac{\mathrm{1}−\mathrm{1}−{x}}{\mathrm{2}{x}\left(\mathrm{1}+{x}\right)} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com