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Question Number 78596 by jagoll last updated on 19/Jan/20

lim_(x→0) (1−3tan^2 x)^((2/(sin^2 3x)) ) = ?

$$ \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \mathrm{x}\right)^{\frac{\mathrm{2}}{\mathrm{sin}\:^{\mathrm{2}} \:\mathrm{3x}}\:\:\:} =\:?\: \\ $$

Commented by mathmax by abdo last updated on 19/Jan/20

let A(x)=(1−3tan^2 x)^(2/(sin^2 (3x))) ⇒ A(x)=e^((2/(sin^2 (3x)))ln(1−3tan^2 x)) we have tanx ∼ x ⇒1−3tan^2 x ∼1−3x^2 and ln(1−3x^2 )∼−3x^2 also sin(3x) ∼3x ⇒sin^2 (3x)∼9x^2 ⇒ (2/(sin^2 (3x)))ln(1−3tan^2 x) ∼((−6x^2 )/(9x^2 )) =−(2/3) ⇒lim_(x→0) A(x)=e^(−(2/3)) =(1/((^3 (√(e^2 )))))

$${let}\:{A}\left({x}\right)=\left(\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}\right)^{\frac{\mathrm{2}}{{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)}} \:\Rightarrow\:{A}\left({x}\right)={e}^{\frac{\mathrm{2}}{{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)}{ln}\left(\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}\right)} \\ $$$${we}\:{have}\:{tanx}\:\sim\:{x}\:\Rightarrow\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}\:\sim\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} \:{and}\:{ln}\left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} \right)\sim−\mathrm{3}{x}^{\mathrm{2}} \\ $$$${also}\:{sin}\left(\mathrm{3}{x}\right)\:\sim\mathrm{3}{x}\:\Rightarrow{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\sim\mathrm{9}{x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\mathrm{2}}{{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)}{ln}\left(\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}\right)\:\sim\frac{−\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{9}{x}^{\mathrm{2}} }\:=−\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {A}\left({x}\right)={e}^{−\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{\left.{e}^{\mathrm{2}} \right)}\right.} \\ $$

Answered by john santu last updated on 19/Jan/20

we know that lim_(x→0) (1+x)^(1/x) = e now we work the question lim_(x→0) [ (1+(−3tan^2 x))^(1/(−3tan^2 x)) ]^((−6tan^2 x)/(sin^2 3x)) = e^(lim_(x→0) (((−6tan^2 x)/(sin^2 3x)))) = e^(−(6/9)) = (1/(e^2 )^(1/(3 )) ) .

$${we}\:{know}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} =\:{e} \\ $$$${now}\:{we}\:{work}\:{the}\:{question}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\left(\mathrm{1}+\left(−\mathrm{3tan}\:^{\mathrm{2}} {x}\right)\right)^{\frac{\mathrm{1}}{−\mathrm{3tan}\:^{\mathrm{2}} {x}}} \right]\:^{\frac{−\mathrm{6tan}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}{x}}} \: \\ $$$$=\:{e}\:^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{−\mathrm{6tan}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}{x}}\right)} \:=\:{e}\:^{−\frac{\mathrm{6}}{\mathrm{9}}} \:=\:\frac{\mathrm{1}}{\sqrt[{\mathrm{3}\:}]{{e}^{\mathrm{2}} }}\:.\: \\ $$

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