Question Number 40118 by maxmathsup by imad last updated on 15/Jul/18
$${calculate}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\mathrm{2}{x}}{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}\:−{cosx} \\ $$
Commented by math khazana by abdo last updated on 19/Jul/18
$${let}\:{A}\left({x}\right)=\:\frac{\mathrm{2}{x}}{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}\:−{cosx} \\ $$$${A}\left({x}\right)\:=\frac{\mathrm{2}{x}}{{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)}\:−{cosx}\:{but} \\ $$$${ln}^{'} \left(\mathrm{1}+{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\mathrm{1}−{x}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\:{and} \\ $$$${ln}\left(\mathrm{1}−{x}\right)\:=−{x}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)=\:\mathrm{2}{x}\:+{o}\left({x}^{\mathrm{3}} \right)\:\:{also} \\ $$$${cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${A}\left({x}\right)\:\:\sim\:\:\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{A}\left({x}\right)=\mathrm{0}\:. \\ $$