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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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