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Question Number 143730 by mathmax by abdo last updated on 17/Jun/21

$$\mathrm{find}{\lim }_{\mathrm{x}\to 0}\frac{\sin \left(\sin (1-\mathrm{cosx})\right)-1+\cos (\mathrm{x}-\mathrm{sinx})}{{\mathrm{x}}^3}$$

Answered by TheHoneyCat last updated on 17/Jun/21

$$\sin x=x-\frac{x^3}{6}+o\left(x^4\right)$$$$x-\sin x=\frac{x^3}{6}+o\left(x^4\right)$$$$\cos x=1-\frac{x^2}{2}+o\left(x^3\right)$$$$\mathrm{thus}:\cos x=1+o\left(x^{12}\right)$$$$\mathrm{so}:-1+\cos x=o\left(x^{12}\right)=o\left(x^3\right)$$$$$$$$1-\cos x=\frac{x^2}{2}+o\left(x^3\right)$$$$\mathrm{so}:\sin (1-\cos x)=\frac{x^2}{2}+o\left(x^3\right)$$$$\mathrm{and}\mathrm{thus}:\sin \left(\sin (1-\cos x)\right)=\frac{x^2}{2}+o\left(x^3\right)$$$$$$$$\mathrm{so}\mathrm{in}\mathrm{the}\mathrm{end}:$$$$\frac{\sin \left(\sin (1-\mathrm{cosx})\right)-1+\cos (\mathrm{x}-\mathrm{sinx})}{{\mathrm{x}}^3}=\frac{1}{2x}+o\left(1\right)$$$$$$$$$$$$$$$${\lim }_{\mathrm{x}\to 0^{+}}\frac{\sin \left(\sin (1-\mathrm{cosx})\right)-1+\cos (\mathrm{x}-\mathrm{sinx})}{{\mathrm{x}}^3}=+\mathrm{\infty }$$$${\lim }_{\mathrm{x}\to 0^{-}}\frac{\sin \left(\sin (1-\mathrm{cosx})\right)-1+\cos (\mathrm{x}-\mathrm{sinx})}{{\mathrm{x}}^3}=-\mathrm{\infty }$$$$$$$$sotechnicalythereisnosuchlimit...$$$$but,weallgetitIhope...$$$$;\bullet )$$

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

$$\mathrm{sinx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{6}\Rightarrow \mathrm{x}-\mathrm{sinx}\sim \frac{{\mathrm{x}}^3}{6}\Rightarrow \cos (\mathrm{x}-\mathrm{sinx})\sim \cos \left(\frac{{\mathrm{x}}^3}{6}\right)\sim 1-\frac{1}{2}\frac{{\mathrm{x}}^6}{36}$$$$=1-\frac{{\mathrm{x}}^6}{72}\Rightarrow \cos (\mathrm{x}-\mathrm{sinx})-1\sim -\frac{{\mathrm{x}}^6}{72}$$$$\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\Rightarrow 1-\mathrm{cosx}\sim \frac{{\mathrm{x}}^2}{2}\Rightarrow \sin (1-\mathrm{cosx})\sim \sin \left(\frac{{\mathrm{x}}^2}{2}\right)\sim \frac{{\mathrm{x}}^2}{2}-\frac{1}{6}{\left(\frac{{\mathrm{x}}^2}{2}\right)}^3$$$$=\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^6}{48}\Rightarrow \sin \left(\sin (1-\mathrm{cosx})\right)\sim \sin (\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^6}{48})\sim \frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^6}{48}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^6}{48}-\frac{{\mathrm{x}}^6}{72}}{{\mathrm{x}}^3}=\frac{1}{2\mathrm{x}}-\frac{{\mathrm{x}}^3}{48}-\frac{{\mathrm{x}}^3}{72}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\infty }$$

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