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Question Number 152247 by mathocean1 last updated on 26/Aug/21

show that lim_(x→0) ((((ln(1+x))/x) −1)/x)=−(1/2)

$${show}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:−\mathrm{1}}{{x}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by mr W last updated on 26/Aug/21

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

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

Commented by peter frank last updated on 27/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mathocean1 last updated on 26/Aug/21

Please sir can you explain or find another way to calculate it? seriously i don′t understand even the first line : i have just finished my studies in high scool...so...

$${Please}\:{sir}\:{can}\:{you}\:{explain}\:{or}\:{find}\: \\ $$$${another}\:{way}\:{to}\:{calculate}\:{it}?\:{seriously} \\ $$$${i}\:{don}'{t}\:{understand}\:{even}\:{the}\:{first}\:{line} \\ $$$$:\:{i}\:{have}\:{just}\:{finished}\:{my}\:{studies}\:{in} \\ $$$${high}\:{scool}...{so}... \\ $$

Commented by mr W last updated on 26/Aug/21

ln (1+x)=x−(x^2 /2)+(x^3 /3)−... see taylor series...

$$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−... \\ $$$${see}\:{taylor}\:{series}... \\ $$

Commented by mr W last updated on 26/Aug/21

https://en.m.wikipedia.org/wiki/Taylor_series

Answered by Mokmokhi last updated on 26/Aug/21

lim((((ln (1+x))/x)−1)/x)=lim((ln (1+x)−x)/x^2 ) by lh rule =lim(((1/(1+x))−1)/(2x)) again =lim((−(1/((1+x)^2 )))/2)=−(1/2)

$${lim}\frac{\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}−\mathrm{1}}{{x}}={lim}\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)−{x}}{{x}^{\mathrm{2}} } \\ $$$${by}\:{lh}\:{rule} \\ $$$$={lim}\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}}−\mathrm{1}}{\mathrm{2}{x}} \\ $$$${again} \\ $$$$={lim}\frac{−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by mr W last updated on 26/Aug/21

using l′hopital rule: lim_(x→0) ((ln (x+1)−x)/x^2 ) =lim_(x→0) (((1/(x+1))−1)/(2x)) =lim_(x→0) ((−1)/(2(x+1))) =−(1/2)

$${using}\:{l}'{hopital}\:{rule}: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)−{x}}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{x}+\mathrm{1}}−\mathrm{1}}{\mathrm{2}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mathocean1 last updated on 26/Aug/21

Thank you very much sirs for your explanations!

$${Thank}\:{you}\:{very}\:{much}\:{sirs}\:{for}\:{your} \\ $$$${explanations}! \\ $$

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