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Question Number 140046 by mnjuly1970 last updated on 03/May/21
........... nice .......calculus(I) ........ Θ :=lim_( x→ (π/4)) ( tan(x) )^( tan(2x)) =? ...............................
$$\:\:\:\:\:\:\:\:\:\:………..\:{nice}\:…….{calculus}\left({I}\right)\:…….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Theta\::={lim}_{\:{x}\rightarrow\:\frac{\pi}{\mathrm{4}}} \:\left(\:{tan}\left({x}\right)\:\right)^{\:{tan}\left(\mathrm{2}{x}\right)} \:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………………………. \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/May/21
thanks alot...
$${thanks}\:{alot}… \\ $$
Answered by Dwaipayan Shikari last updated on 03/May/21
lim_(z→0) tan((π/2)+2z)log(tan((π/4)+z))=log(y) −cot(2z)log(((1+tanz)/(1−tanz)))=log(y) ∼−cot(2z)log(1+z)+cot(2z)log(1−z)=log(y) ≈((−1)/(2z))z+((−z)/(2z))=log(y)⇒log(y)=−1⇒y=(1/e)
$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}{tan}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{z}\right){log}\left({tan}\left(\frac{\pi}{\mathrm{4}}+{z}\right)\right)={log}\left({y}\right) \\ $$$$−{cot}\left(\mathrm{2}{z}\right){log}\left(\frac{\mathrm{1}+{tanz}}{\mathrm{1}−{tanz}}\right)={log}\left({y}\right) \\ $$$$\sim−{cot}\left(\mathrm{2}{z}\right){log}\left(\mathrm{1}+{z}\right)+{cot}\left(\mathrm{2}{z}\right){log}\left(\mathrm{1}−{z}\right)={log}\left({y}\right) \\ $$$$\approx\frac{−\mathrm{1}}{\mathrm{2}{z}}{z}+\frac{−{z}}{\mathrm{2}{z}}={log}\left({y}\right)\Rightarrow{log}\left({y}\right)=−\mathrm{1}\Rightarrow{y}=\frac{\mathrm{1}}{{e}} \\ $$
Answered by mnjuly1970 last updated on 04/May/21
solution..... tan(x):=y ⇒ {_(y→1) ^(x→(π/4)) Θ:=lim_(y→1) (1−(1−y))^((2y)/((1−y)(1+y))) :=^(1−y=t) lim_(t→0) (1−t)^((2(1−t))/(t(2−t))) =lim_(t→0) {(1−t)^(2/t) }^((1−t)/(2−t)) :=(e^(−2) )^(1/2) =e^(−1) =(1/e) ......✓✓ .........Θ :=(1/e) ........
$$\:\:\:\:\:\:{solution}….. \\ $$$$\:\:\:\:\:\:\:{tan}\left({x}\right):={y}\:\Rightarrow\:\left\{_{{y}\rightarrow\mathrm{1}} ^{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \right. \\ $$$$\:\:\:\:\:\:\:\Theta:={lim}_{{y}\rightarrow\mathrm{1}} \left(\mathrm{1}−\left(\mathrm{1}−{y}\right)\right)^{\frac{\mathrm{2}{y}}{\left(\mathrm{1}−{y}\right)\left(\mathrm{1}+{y}\right)}} \\ $$$$\:\:\:\:\::\overset{\mathrm{1}−{y}={t}} {=}{lim}_{{t}\rightarrow\mathrm{0}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{2}\left(\mathrm{1}−{t}\right)}{{t}\left(\mathrm{2}−{t}\right)}} ={lim}_{{t}\rightarrow\mathrm{0}} \left\{\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{2}}{{t}}} \right\}^{\frac{\mathrm{1}−{t}}{\mathrm{2}−{t}}} \\ $$$$\:\:\:\:\:\::=\left({e}^{−\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={e}^{−\mathrm{1}} =\frac{\mathrm{1}}{{e}}\:……\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………\Theta\::=\frac{\mathrm{1}}{{e}}\:…….. \\ $$

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