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Question Number 93033 by i jagooll last updated on 10/May/20

lim_(x→π/6) (tan (((3x)/2)))^(tan (3x)) =?

$$\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\:\left(\mathrm{tan}\:\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\right)^{\mathrm{tan}\:\left(\mathrm{3x}\right)} =?\: \\ $$

Commented by john santu last updated on 10/May/20

let ((3x)/2) = t ⇒3x = 2t with t → (π/4) lim_(t→π/4) (tan t)^(tan 2t) = lim_(t→π/4) ln y = lim_(t→π/4) tan 2t ln (tan t) lim_(t→π/4) ln y = lim_(t→π/4) ((ln (tan t))/(cot 2t)) = lim_(t→π/4) ((sec^2 t)/(tan t)) × (−((sin^2 2t)/2)) = −(1/2)× (2/1) = −1 . then lim_(x→3π/6) (tan (((3x)/2)))^(tan (3x)) = e^(−1) = (1/e)

$$\mathrm{let}\:\frac{\mathrm{3x}}{\mathrm{2}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{3x}\:=\:\mathrm{2t}\: \\ $$$$\mathrm{with}\:\mathrm{t}\:\rightarrow\:\frac{\pi}{\mathrm{4}} \\ $$$$\underset{\mathrm{t}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\left(\mathrm{tan}\:\mathrm{t}\right)^{\mathrm{tan}\:\mathrm{2t}} \:=\: \\ $$$$\underset{\mathrm{t}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\mathrm{ln}\:\mathrm{y}\:=\:\underset{\mathrm{t}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\mathrm{tan}\:\mathrm{2t}\:\mathrm{ln}\:\left(\mathrm{tan}\:\mathrm{t}\right)\: \\ $$$$\underset{\mathrm{t}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\mathrm{ln}\:\mathrm{y}\:=\:\underset{\mathrm{t}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{tan}\:\mathrm{t}\right)}{\mathrm{cot}\:\mathrm{2t}} \\ $$$$=\:\underset{\mathrm{t}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{t}}{\mathrm{tan}\:\mathrm{t}}\:×\:\left(−\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{2t}}{\mathrm{2}}\right) \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}×\:\frac{\mathrm{2}}{\mathrm{1}}\:=\:−\mathrm{1}\:.\: \\ $$$$\mathrm{then}\:\underset{{x}\rightarrow\mathrm{3}\pi/\mathrm{6}} {\mathrm{lim}}\:\left(\mathrm{tan}\:\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\right)^{\mathrm{tan}\:\left(\mathrm{3x}\right)} \\ $$$$=\:\mathrm{e}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{e}}\: \\ $$

Commented by i jagooll last updated on 10/May/20

good sir...

Commented by mathmax by abdo last updated on 11/May/20

let f(x)=(tan(((3x)/2))^(tan(3x)) ⇒ln(f(x)) =tan(3x)ln(tan(((3x)/2))) changement x =−t+(π/6) give ln(f(x)) =tan(3(−t+(π/6)))ln(tan((3/2)(−t+(π/6))) =tan((π/2)−3t)ln(tan((π/4)−((3t)/2)))=((ln(tan((π/4)−((3t)/2))))/(tan(3t))) =((ln(((1−tan(((3t)/2)))/(1+tan(((3t)/2))))))/(tan(3t))) ∼(1/(3t))ln(((1−((3t)/2))/(((3t)/2)+1)))=(1/(3t)){ln(1−((3t)/2))−ln(1+((3t)/2))} ∼(1/(3t)){−((3t)/2)−((3t)/2)} =−1 ⇒lim(f(x)) =−1 ⇒lim_(x→(π/6)) f(x) =e^(−1) =(1/e)

$${let}\:{f}\left({x}\right)=\left({tan}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)^{{tan}\left(\mathrm{3}{x}\right)} \:\Rightarrow{ln}\left({f}\left({x}\right)\right)\:={tan}\left(\mathrm{3}{x}\right){ln}\left({tan}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\right)\right. \\ $$$${changement}\:{x}\:=−{t}+\frac{\pi}{\mathrm{6}}\:{give}\:{ln}\left({f}\left({x}\right)\right)\:={tan}\left(\mathrm{3}\left(−{t}+\frac{\pi}{\mathrm{6}}\right)\right){ln}\left({tan}\left(\frac{\mathrm{3}}{\mathrm{2}}\left(−{t}+\frac{\pi}{\mathrm{6}}\right)\right)\right. \\ $$$$={tan}\left(\frac{\pi}{\mathrm{2}}−\mathrm{3}{t}\right){ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\mathrm{3}{t}}{\mathrm{2}}\right)\right)=\frac{{ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\mathrm{3}{t}}{\mathrm{2}}\right)\right)}{{tan}\left(\mathrm{3}{t}\right)} \\ $$$$=\frac{{ln}\left(\frac{\mathrm{1}−{tan}\left(\frac{\mathrm{3}{t}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}\left(\frac{\mathrm{3}{t}}{\mathrm{2}}\right)}\right)}{{tan}\left(\mathrm{3}{t}\right)}\:\sim\frac{\mathrm{1}}{\mathrm{3}{t}}{ln}\left(\frac{\mathrm{1}−\frac{\mathrm{3}{t}}{\mathrm{2}}}{\frac{\mathrm{3}{t}}{\mathrm{2}}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{3}{t}}\left\{{ln}\left(\mathrm{1}−\frac{\mathrm{3}{t}}{\mathrm{2}}\right)−{ln}\left(\mathrm{1}+\frac{\mathrm{3}{t}}{\mathrm{2}}\right)\right\} \\ $$$$\sim\frac{\mathrm{1}}{\mathrm{3}{t}}\left\{−\frac{\mathrm{3}{t}}{\mathrm{2}}−\frac{\mathrm{3}{t}}{\mathrm{2}}\right\}\:=−\mathrm{1}\:\Rightarrow{lim}\left({f}\left({x}\right)\right)\:=−\mathrm{1}\:\Rightarrow{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {f}\left({x}\right)\:={e}^{−\mathrm{1}} \:=\frac{\mathrm{1}}{{e}} \\ $$

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