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Question Number 99505 by bemath last updated on 21/Jun/20

Commented by Dwaipayan Shikari last updated on 21/Jun/20

Ans:1

$${Ans}:\mathrm{1} \\ $$

Commented by Dwaipayan Shikari last updated on 21/Jun/20

=e^(lim_(x→(π/2)) (sinx)(1−tan((x/2)))) =e^(lim_(x→(π/2)) (sinx−2sin^2 ((x/2)))) =e^((1−1)) =e^0 =1

$$={e}^{{li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{m}}\left({sinx}\right)\left(\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)} \\ $$$$={e}^{{li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{m}}\left({sinx}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)} \\ $$$$={e}^{\left(\mathrm{1}−\mathrm{1}\right)} ={e}^{\mathrm{0}} =\mathrm{1} \\ $$

Answered by john santu last updated on 22/Jun/20

let tan ((x/2)) = t lim_(t→1) (((1−2t+t^2 )/(1+t^2 ))) ^(t−1) = e^(lim_(t→1) ln (((1−2t+t^2 )/(1+t^2 )))^(t−1) ) e^(lim_(t→1) ((ln(((1−2t+t^2 )/(1+t^2 ))))/(t−1))) = e^(lim_(t→1) (((((1+t^2 )/((t−1)^2 )))((((2t−2)(1+t^2 )−2t(t−1)^2 )/((1+t^2 )^2 ))))/1)) = e^0 = 1

$$\mathrm{let}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:=\:\mathrm{t} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−\mathrm{2t}+\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)\overset{\mathrm{t}−\mathrm{1}} {\:}\:=\:\mathrm{e}^{\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{ln}\:\left(\frac{\mathrm{1}−\mathrm{2t}+\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)^{\mathrm{t}−\mathrm{1}} } \\ $$$$\mathrm{e}^{\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\frac{\mathrm{1}−\mathrm{2t}+\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)}{\mathrm{t}−\mathrm{1}}} =\:\mathrm{e}^{\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }\right)\left(\frac{\left(\mathrm{2t}−\mathrm{2}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)−\mathrm{2t}\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)}{\mathrm{1}}} \\ $$$$=\:\mathrm{e}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 21/Jun/20

let f(x) =(1−sinx)^(tan((x/2)−1)) ⇒f(x) =e^((tan((x/2))−1)ln(1−sinx)) changement x =(π/2)−t give f(x) =e^((tan((π/4)−(t/2))−1)ln(1−cost)) =g(t) (t→0) we have tan((π/4)−(t/2))−1 =((1−tan((t/2)))/(1+tan((t/2))))−1 =((−2tan((t/2)))/(1+tan((t/2)))) ∼((−t)/(1+(t/2)))∼−t 1−cost ∼(t^2 /2) ⇒ln(1−cost) ∼2lnt−ln(2) ⇒ g(t) ∼e^(−t(2lnt−ln2)) →1 (t→0) ⇒lim_(x→(π/2)) f(x) =1

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{1}−\mathrm{sinx}\right)^{\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{1}\right)} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\left(\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}−\mathrm{sinx}\right)} \\ $$$$\mathrm{changement}\:\mathrm{x}\:=\frac{\pi}{\mathrm{2}}−\mathrm{t}\:\:\mathrm{give}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\left(\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\mathrm{t}}{\mathrm{2}}\right)−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}−\mathrm{cost}\right)} \:=\mathrm{g}\left(\mathrm{t}\right) \\ $$$$\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\mathrm{t}}{\mathrm{2}}\right)−\mathrm{1}\:=\frac{\mathrm{1}−\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}−\mathrm{1}\:=\frac{−\mathrm{2tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}\:\sim\frac{−\mathrm{t}}{\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}}\sim−\mathrm{t} \\ $$$$\mathrm{1}−\mathrm{cost}\:\sim\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{cost}\right)\:\sim\mathrm{2lnt}−\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{t}\right)\:\sim\mathrm{e}^{−\mathrm{t}\left(\mathrm{2lnt}−\mathrm{ln2}\right)} \:\:\rightarrow\mathrm{1}\:\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{1} \\ $$

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