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Question Number 29459 by prof Abdo imad last updated on 08/Feb/18

find lim_(x→0)    (((sinx)/x))^(1/(1−cosx))  .

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{1}−{cosx}}} \:. \\ $$

Commented by Cheyboy last updated on 09/Feb/18

(1/(e)^(1/3) )

$$\frac{\mathrm{1}}{\sqrt[{\mathrm{3}}]{{e}}} \\ $$

Commented by mrW2 last updated on 09/Feb/18

to cheyboy sir:  sir, it would make more sense, if you  could also show your working instead  of only a result. thank you!

$${to}\:{cheyboy}\:{sir}: \\ $$$${sir},\:{it}\:{would}\:{make}\:{more}\:{sense},\:{if}\:{you} \\ $$$${could}\:{also}\:{show}\:{your}\:{working}\:{instead} \\ $$$${of}\:{only}\:{a}\:{result}.\:{thank}\:{you}! \\ $$

Commented by Cheyboy last updated on 09/Feb/18

sir i was having low battery thats  why i could not type the working.  thanks for that sir i will stick to  your word next time.

$${sir}\:{i}\:{was}\:{having}\:{low}\:{battery}\:{thats} \\ $$$${why}\:{i}\:{could}\:{not}\:{type}\:{the}\:{working}. \\ $$$${thanks}\:{for}\:{that}\:{sir}\:{i}\:{will}\:{stick}\:{to} \\ $$$${your}\:{word}\:{next}\:{time}. \\ $$

Commented by mrW2 last updated on 09/Feb/18

thanks!

$${thanks}! \\ $$

Commented by prof Abdo imad last updated on 10/Feb/18

let put A(x)= (((sinx)/x))^(1/(1−cosx)) ⇒ln(A(x))=  (1/(1−cosx))ln(((sinx)/x))  sinx =x −(x^3 /(3!)) +o(x^5 )⇒ ((sinx)/x) =1− (x^2 /7) +o(x^4 )and  ln(((sinx)/x))=ln(1−(x^2 /6) +o(x^4 )) ∼ −(x^2 /6) we have also  cos x ∼1 −(x^2 /2) ⇒1−cosx∼ (x^2 /2) so  (1/(1−cosx))ln(((sinx)/x))  ∼ (2/x^2 )×(−(x^2 /6))= −(1/3) ⇒  lim_(x→0) ln(A(x))=−(1/3) ⇒lim_(x→0) A(x)=e^(−(1/3))   = (^3 (√e))^(−1) .

$${let}\:{put}\:{A}\left({x}\right)=\:\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{1}−{cosx}}} \Rightarrow{ln}\left({A}\left({x}\right)\right)=\:\:\frac{\mathrm{1}}{\mathrm{1}−{cosx}}{ln}\left(\frac{{sinx}}{{x}}\right) \\ $$$${sinx}\:={x}\:−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+{o}\left({x}^{\mathrm{5}} \right)\Rightarrow\:\frac{{sinx}}{{x}}\:=\mathrm{1}−\:\frac{{x}^{\mathrm{2}} }{\mathrm{7}}\:+{o}\left({x}^{\mathrm{4}} \right){and} \\ $$$${ln}\left(\frac{{sinx}}{{x}}\right)={ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\:+{o}\left({x}^{\mathrm{4}} \right)\right)\:\sim\:−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\:{we}\:{have}\:{also} \\ $$$${cos}\:{x}\:\sim\mathrm{1}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{1}−{cosx}\sim\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{so} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{cosx}}{ln}\left(\frac{{sinx}}{{x}}\right)\:\:\sim\:\frac{\mathrm{2}}{{x}^{\mathrm{2}} }×\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\right)=\:−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} {ln}\left({A}\left({x}\right)\right)=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {A}\left({x}\right)={e}^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\:\left(\:^{\mathrm{3}} \sqrt{{e}}\right)^{−\mathrm{1}} . \\ $$$$ \\ $$

Commented by ajfour last updated on 10/Feb/18

thanks for solution, Sir .

$${thanks}\:{for}\:{solution},\:{Sir}\:. \\ $$

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