Question-205379

Question Number 205379 by cortano12 last updated on 19/Mar/24

Answered by MM42 last updated on 19/Mar/24

hop→=lim_(x→0) ((−cosx×sin(sinx)+sinx)/(4x^3 )) hop→=((sinx×sin(sinx)−cos^2 x×cos(sinx)+cosx)/(12x^2 )) =lim_(x→0) ((cosx(1−cosx))/(12x^2 )) =lim_(x→0) (((1/2)x^2 )/(12))=(1/(24)) ✓

$${hop}\rightarrow={lim}_{{x}\rightarrow\mathrm{0}} \frac{−{cosx}×{sin}\left({sinx}\right)+{sinx}}{\mathrm{4}{x}^{\mathrm{3}} } \\ $$$${hop}\rightarrow=\frac{{sinx}×{sin}\left({sinx}\right)−{cos}^{\mathrm{2}} {x}×{cos}\left({sinx}\right)+{cosx}}{\mathrm{12}{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{cosx}\left(\mathrm{1}−{cosx}\right)}{\mathrm{12}{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{24}}\:\:\checkmark \\ $$$$ \\ $$

Commented by Frix last updated on 20/Mar/24

$$\mathrm{I}\:\mathrm{get}\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Answered by namphamduc last updated on 22/Mar/24

L=lim_(x→0) ((cos(sin(x))−cos(x))/x^4 )=−2lim_(x→0) ((sin(((sin(x)+x)/2))sin(((sin(x)−x)/2)))/x^4 ) =−2lim_(x→0) ((sin(((sin(x)+x)/2)))/((sin(x)+x)/2))∙((sin(((sin(x)−x)/2)))/((sin(x)−x)/2)).((sin(x)+x)/(2x)).((sin(x)−x)/(2x^3 )) lim_(x→0) ((sin(x)−x)/x^3 )=−(1/6) is well−known ⇒L=−2.1.1.((1+1)/2)(−(1/6))(1/2)=(1/6)

$$ \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\left(\mathrm{sin}\left({x}\right)\right)−\mathrm{cos}\left({x}\right)}{{x}^{\mathrm{4}} }=−\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{sin}\left({x}\right)+{x}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\mathrm{sin}\left({x}\right)−{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{4}} } \\ $$$$=−\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{sin}\left({x}\right)+{x}}{\mathrm{2}}\right)}{\frac{\mathrm{sin}\left({x}\right)+{x}}{\mathrm{2}}}\centerdot\frac{\mathrm{sin}\left(\frac{\mathrm{sin}\left({x}\right)−{x}}{\mathrm{2}}\right)}{\frac{\mathrm{sin}\left({x}\right)−{x}}{\mathrm{2}}}.\frac{\mathrm{sin}\left({x}\right)+{x}}{\mathrm{2}{x}}.\frac{\mathrm{sin}\left({x}\right)−{x}}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left({x}\right)−{x}}{{x}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{is}\:\mathrm{well}−\mathrm{known} \\ $$$$\Rightarrow{L}=−\mathrm{2}.\mathrm{1}.\mathrm{1}.\frac{\mathrm{1}+\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{6}}\right)\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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