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Question Number 68178 by Mikael last updated on 06/Sep/19

lim_(x→0) (((cosx)^(sin2x) −1)/x^3 )=?

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

Commented by mathmax by abdo last updated on 06/Sep/19

let f(x) =(((cosx)^(sin(2x)) −1)/x^3 ) we have f(x)=((e^(sin(2x)ln(cosx)) −1)/x^3 ) we have cosx ∼1−(x^2 /2) ⇒ln(cosx)∼ln(1−(x^2 /2)) ln^′ (1−u) =((−1)/(1−u)) =−(1+u+o(u)) ⇒ ln(1−u) =−u−(u^2 /2) +o(u^2 ) ⇒ln(1−(x^2 /2))=−(x^2 /2)−(x^4 /4) +o(x^4 ) also sin(2x)∼2x⇒sin(2x)ln(cosx) ∼2x(−(x^2 /2)−(x^4 /4)) =−x^3 −(1/2)x^5 ⇒e^(sin(2x)ln(cosx)) ∼e^(−x^3 −(1/2)x^5 ) ∼1−x^3 −(1/2)x^5 ⇒ f(x)∼ ((−x^3 −(1/2)x^5 )/x^3 ) ⇒f(x)∼−1−(1/2)x^2 ⇒lim_(x→0) f(x) =−1

$${let}\:{f}\left({x}\right)\:=\frac{\left({cosx}\right)^{{sin}\left(\mathrm{2}{x}\right)} −\mathrm{1}}{{x}^{\mathrm{3}} }\:\:{we}\:{have}\:{f}\left({x}\right)=\frac{{e}^{{sin}\left(\mathrm{2}{x}\right){ln}\left({cosx}\right)} −\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$${we}\:{have}\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{ln}\left({cosx}\right)\sim{ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$${ln}^{'} \left(\mathrm{1}−{u}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}−{u}}\:=−\left(\mathrm{1}+{u}+{o}\left({u}\right)\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{u}\right)\:=−{u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow{ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\:+{o}\left({x}^{\mathrm{4}} \right) \\ $$$${also}\:\:{sin}\left(\mathrm{2}{x}\right)\sim\mathrm{2}{x}\Rightarrow{sin}\left(\mathrm{2}{x}\right){ln}\left({cosx}\right)\:\sim\mathrm{2}{x}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right) \\ $$$$=−{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{5}} \:\Rightarrow{e}^{{sin}\left(\mathrm{2}{x}\right){ln}\left({cosx}\right)} \sim{e}^{−{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{5}} } \:\sim\mathrm{1}−{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{5}} \:\Rightarrow \\ $$$${f}\left({x}\right)\sim\:\:\frac{−{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{5}} }{{x}^{\mathrm{3}} }\:\Rightarrow{f}\left({x}\right)\sim−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:=−\mathrm{1} \\ $$

Commented by Mikael last updated on 06/Sep/19

thank you.

$${thank}\:{you}. \\ $$

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