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Question Number 29846 by abdo imad last updated on 12/Feb/18
give the developpement at integr series for f(x)=((ln(1+x)−ln(1−x))/x) 2)find lim_(x→0) f(x).
$${give}\:{the}\:{developpement}\:\:{at}\:{integr}\:{series}\:{for} \\ $$$${f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)}{{x}} \\ $$$$\left.\mathrm{2}\right){find}\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right). \\ $$
Commented by maxmathsup by imad last updated on 09/Apr/19
1) we have (d/dx)(ln(1+x)) =(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ln(1+x) =Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) let change x by −x ⇒ ln(1−x) =Σ_(n=1) ^∞ (((−1)^(n−1) (−x)^n )/n) =−Σ_(n=1) ^∞ (x^n /n) ⇒ ln(1+x)−ln(1−x) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) +Σ_(n=1) ^∞ (x^n /n) =Σ_(n=1) ^∞ (((1−(−1)^n )/n))x^n =Σ_(n=1) ^∞ (2/(2n+1)) x^(2n+1) ⇒f(x) =2 Σ_(n=1) ^∞ (x^(2n) /(2n+1)) with ∣x∣<1 and x≠0 2) we have ln(1+x) ∼ x (x ∈v(0)) and ln(1−x)∼−x ⇒ ln(1+x)−ln(1−x) ∼2x ⇒((ln(1+x)−ln(1−x))/x) ∼ 2 ⇒ lim_(x→0) f(x) =2 .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\frac{{d}}{{dx}}\left({ln}\left(\mathrm{1}+{x}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}} \:\:\Rightarrow{ln}\left(\mathrm{1}+{x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{x}^{{n}} }{{n}}\:\:\:{let}\:{change}\:{x}\:{by}\:−{x}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} \:\left(−\boldsymbol{{x}}\right)^{\boldsymbol{{n}}} }{\boldsymbol{{n}}}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{x}^{{n}} }{{n}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} }{{n}}\right){x}^{{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:\Rightarrow{f}\left({x}\right)\:=\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$${with}\:\mid{x}\mid<\mathrm{1}\:\:\:{and}\:{x}\neq\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\:{we}\:{have}\:{ln}\left(\mathrm{1}+{x}\right)\:\sim\:{x}\:\:\:\left({x}\:\in{v}\left(\mathrm{0}\right)\right)\:\:{and}\:{ln}\left(\mathrm{1}−{x}\right)\sim−{x}\:\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)\:\sim\mathrm{2}{x}\:\Rightarrow\frac{{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:\sim\:\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)\:=\mathrm{2}\:. \\ $$

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