Question Number 59187 by maxmathsup by imad last updated on 05/May/19
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left({arctan}\left(\mathrm{1}+{x}\right)\right)−{ln}\left(\frac{\pi}{\mathrm{4}}\right)}{{x}^{\mathrm{2}} } \\ $$
Commented by kaivan.ahmadi last updated on 05/May/19
$${hop} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}{xarctg}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \right)}=\infty \\ $$
Commented by maxmathsup by imad last updated on 06/May/19
$${let}\:{use}\:{hospital}\:{theorem}\:\:{let}\:{f}\left({x}\right)={ln}\left({arctan}\left(\mathrm{1}+{x}\right)\right)−{ln}\left(\frac{\pi}{\mathrm{4}}\right)\:{and} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} \:\:\:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }}{{arctan}\left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{\left(\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \right){arctan}\left(\mathrm{1}+{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}+\mathrm{1}\right)\:{arctan}\left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right){arctan}\left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{{u}\left({x}\right)}\:\Rightarrow \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−\frac{{u}^{'} \left({x}\right)}{\left({u}\left({x}\right)\right)^{\mathrm{2}} }\:\:{but}\:\:{u}^{'} \left({x}\right)\:=\left(\mathrm{2}{x}+\mathrm{2}\right){arctan}\left(\mathrm{1}+{x}\right)\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$$$=\left(\mathrm{2}{x}+\mathrm{2}\right)\:{arctan}\left(\mathrm{1}+{x}\right)\:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−\frac{\left(\mathrm{2}{x}+\mathrm{2}\right){arctan}\left(\mathrm{1}+{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} \left({arctan}\left(\mathrm{1}+{x}\right)\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\frac{\mathrm{2}.\frac{\pi}{\mathrm{4}}}{\mathrm{4}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}.\frac{\pi}{\mathrm{4}}}\:=\frac{\mathrm{2}}{\pi}\:\:\:{also}\:{we}\:{have}\:{g}^{'} \left({x}\right)=\mathrm{2}{x}\:{and}\:{g}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left({arctan}\left(\mathrm{1}+{x}\right)\right)−{ln}\left(\frac{\pi}{\mathrm{4}}\right)}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\pi}\:. \\ $$$$ \\ $$