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Question Number 66321 by mathmax by abdo last updated on 12/Aug/19

find lim_(x→0^+ ) (tan((π/(2+x))))^x

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)^{{x}} \\ $$

Commented by mathmax by abdo last updated on 24/Aug/19

let A(x) =(tan((π/(2+x))))^x ⇒ A(x) =e^(xln((π/(2+x)))) changement (π/(2+x)) =t give ((2+x)/π) =(1/t) ⇒2+x =(π/t) ⇒x =(π/t)−2 ⇒ A(x) =B(t) =e^(((π/t)−2)ln(tant)) ⇒lim_(x→0) A(x)=lim_(t→(π/2)) e^(((π/t)−2)ln(t)) =lim_(t→(π/2)) e^((π−2t)((ln(tant))/(2t))) changement π−2t =u give 2t=π−u lim_(t→(π/2)) B(t) =lim_(u→0) e^(u((ln(tan((π/2)−(u/2))))/(π−u))) =lim_(u→0) e^((u/(π−u))ln((1/(tan((u/2)))))) =lim_(u→0) e^(−(u/(π−u))ln(tan((u/2)))) =1 because tan((u/2))∼(u/2)(V(0)) and lim_(u→0^+ ) ulnu=0

$${let}\:{A}\left({x}\right)\:=\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)^{{x}} \:\Rightarrow\:{A}\left({x}\right)\:={e}^{{xln}\left(\frac{\pi}{\mathrm{2}+{x}}\right)} \\ $$$${changement}\:\frac{\pi}{\mathrm{2}+{x}}\:={t}\:{give}\:\frac{\mathrm{2}+{x}}{\pi}\:=\frac{\mathrm{1}}{{t}}\:\Rightarrow\mathrm{2}+{x}\:=\frac{\pi}{{t}}\:\Rightarrow{x}\:=\frac{\pi}{{t}}−\mathrm{2}\:\Rightarrow \\ $$$${A}\left({x}\right)\:={B}\left({t}\right)\:={e}^{\left(\frac{\pi}{{t}}−\mathrm{2}\right){ln}\left({tant}\right)} \:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)={lim}_{{t}\rightarrow\frac{\pi}{\mathrm{2}}} \:{e}^{\left(\frac{\pi}{{t}}−\mathrm{2}\right){ln}\left({t}\right)} \\ $$$$={lim}_{{t}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:{e}^{\left(\pi−\mathrm{2}{t}\right)\frac{{ln}\left({tant}\right)}{\mathrm{2}{t}}} \:\:{changement}\:\pi−\mathrm{2}{t}\:={u}\:{give}\:\mathrm{2}{t}=\pi−{u} \\ $$$${lim}_{{t}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:{B}\left({t}\right)\:={lim}_{{u}\rightarrow\mathrm{0}} \:\:\:{e}^{{u}\frac{{ln}\left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{{u}}{\mathrm{2}}\right)\right)}{\pi−{u}}} \\ $$$$={lim}_{{u}\rightarrow\mathrm{0}} \:\:\:\:{e}^{\frac{{u}}{\pi−{u}}{ln}\left(\frac{\mathrm{1}}{{tan}\left(\frac{{u}}{\mathrm{2}}\right)}\right)} \:={lim}_{{u}\rightarrow\mathrm{0}} \:\:\:{e}^{−\frac{{u}}{\pi−{u}}{ln}\left({tan}\left(\frac{{u}}{\mathrm{2}}\right)\right)} \:=\mathrm{1}\: \\ $$$${because}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)\sim\frac{{u}}{\mathrm{2}}\left({V}\left(\mathrm{0}\right)\right)\:{and}\:{lim}_{{u}\rightarrow\mathrm{0}^{+} } \:{ulnu}=\mathrm{0} \\ $$

Commented by mathmax by abdo last updated on 24/Aug/19

another way we have (π/(2+x)) =(π/(2(1+(x/2)))) ∼(π/2)(1−(x/2))=(π/2)−((πx)/4) ⇒tan((π/(2+x))) =(1/(tan(((πx)/4)))) but (tan((π/(2+x))))^x =e^(xln(tan((π/(2+x))))) ⇒f(x) =e^(−xln(tan(((πx)/4)))) ∼ e^(−xln(((πx)/4))) →1 (x→0)

$${another}\:{way}\:\:{we}\:{have}\:\frac{\pi}{\mathrm{2}+{x}}\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)}\:\sim\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi{x}}{\mathrm{4}} \\ $$$$\Rightarrow{tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\:=\frac{\mathrm{1}}{{tan}\left(\frac{\pi{x}}{\mathrm{4}}\right)}\:\:{but}\:\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)^{{x}} \:={e}^{{xln}\left({tan}\left(\frac{\pi}{\mathrm{2}+{x}}\right)\right)} \\ $$$$\Rightarrow{f}\left({x}\right)\:\:={e}^{−{xln}\left({tan}\left(\frac{\pi{x}}{\mathrm{4}}\right)\right)} \:\sim\:{e}^{−{xln}\left(\frac{\pi{x}}{\mathrm{4}}\right)} \:\rightarrow\mathrm{1}\:\:\left({x}\rightarrow\mathrm{0}\right) \\ $$

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