Question Number 66321 by mathmax by abdo last updated on 12/Aug/19
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)^{{x}} \\ $$
Commented by mathmax by abdo last updated on 24/Aug/19
$${let}\:{A}\left({x}\right)\:=\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)^{{x}} \:\Rightarrow\:{A}\left({x}\right)\:={e}^{{xln}\left(\frac{\pi}{\mathrm{2}+{x}}\right)} \\ $$$${changement}\:\frac{\pi}{\mathrm{2}+{x}}\:={t}\:{give}\:\frac{\mathrm{2}+{x}}{\pi}\:=\frac{\mathrm{1}}{{t}}\:\Rightarrow\mathrm{2}+{x}\:=\frac{\pi}{{t}}\:\Rightarrow{x}\:=\frac{\pi}{{t}}−\mathrm{2}\:\Rightarrow \\ $$$${A}\left({x}\right)\:={B}\left({t}\right)\:={e}^{\left(\frac{\pi}{{t}}−\mathrm{2}\right){ln}\left({tant}\right)} \:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)={lim}_{{t}\rightarrow\frac{\pi}{\mathrm{2}}} \:{e}^{\left(\frac{\pi}{{t}}−\mathrm{2}\right){ln}\left({t}\right)} \\ $$$$={lim}_{{t}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:{e}^{\left(\pi−\mathrm{2}{t}\right)\frac{{ln}\left({tant}\right)}{\mathrm{2}{t}}} \:\:{changement}\:\pi−\mathrm{2}{t}\:={u}\:{give}\:\mathrm{2}{t}=\pi−{u} \\ $$$${lim}_{{t}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:{B}\left({t}\right)\:={lim}_{{u}\rightarrow\mathrm{0}} \:\:\:{e}^{{u}\frac{{ln}\left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{{u}}{\mathrm{2}}\right)\right)}{\pi−{u}}} \\ $$$$={lim}_{{u}\rightarrow\mathrm{0}} \:\:\:\:{e}^{\frac{{u}}{\pi−{u}}{ln}\left(\frac{\mathrm{1}}{{tan}\left(\frac{{u}}{\mathrm{2}}\right)}\right)} \:={lim}_{{u}\rightarrow\mathrm{0}} \:\:\:{e}^{−\frac{{u}}{\pi−{u}}{ln}\left({tan}\left(\frac{{u}}{\mathrm{2}}\right)\right)} \:=\mathrm{1}\: \\ $$$${because}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)\sim\frac{{u}}{\mathrm{2}}\left({V}\left(\mathrm{0}\right)\right)\:{and}\:{lim}_{{u}\rightarrow\mathrm{0}^{+} } \:{ulnu}=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 24/Aug/19
$${another}\:{way}\:\:{we}\:{have}\:\frac{\pi}{\mathrm{2}+{x}}\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)}\:\sim\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi{x}}{\mathrm{4}} \\ $$$$\Rightarrow{tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\:=\frac{\mathrm{1}}{{tan}\left(\frac{\pi{x}}{\mathrm{4}}\right)}\:\:{but}\:\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)^{{x}} \:={e}^{{xln}\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)} \\ $$$$\Rightarrow{f}\left({x}\right)\:\:={e}^{−{xln}\left({tan}\left(\frac{\pi{x}}{\mathrm{4}}\right)\right)} \:\sim\:{e}^{−{xln}\left(\frac{\pi{x}}{\mathrm{4}}\right)} \:\rightarrow\mathrm{1}\:\:\left({x}\rightarrow\mathrm{0}\right) \\ $$