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Question Number 110449 by mathmax by abdo last updated on 29/Aug/20

$$\mathrm{find}{\lim }_{\mathrm{x}\to 0}\frac{\arctan (\mathrm{x}-\mathrm{sinx})-\arctan (1-\mathrm{cosx})}{{\mathrm{x}}^2}$$

Answered by bemath last updated on 29/Aug/20

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}-(1-\cos \mathrm{x})}{{\mathrm{x}}^2}=$$$$\underset{x\to 0}{\lim }\frac{\mathrm{x}-(\mathrm{x}-\frac{1}{6}{\mathrm{x}}^3)-(1-(1-\frac{1}{2}{\mathrm{x}}^2))}{{\mathrm{x}}^2}=$$$$\underset{x\to 0}{\lim }\frac{\frac{1}{6}{\mathrm{x}}^3-\frac{1}{2}{\mathrm{x}}^2}{{\mathrm{x}}^2}=-\frac{1}{2}$$

Answered by mathmax by abdo last updated on 29/Aug/20

$$\mathrm{let}\mathrm{use}\mathrm{hospital}\mathrm{theorem}\mathrm{u}\left(\mathrm{x}\right)=\arctan (\mathrm{x}-\mathrm{sinx})-\arctan (1-\mathrm{cosx})$$$$\mathrm{v}\left(\mathrm{x}\right)={\mathrm{x}}^2\mathrm{we}\mathrm{have}{\mathrm{u}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1-\mathrm{cosx}}{1+{(\mathrm{x}-\mathrm{sinx})}^2}-\frac{\mathrm{sinx}}{1+{(1-\mathrm{cosx})}^2}$$$$\Rightarrow {\mathrm{u}}^{\left(2\right)}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}(1+{(\mathrm{x}-\mathrm{sinx})}^2)-2(1-\mathrm{cosx})(1-\mathrm{cosx})}{\{1+{(\mathrm{x}-\mathrm{sinx}\}}^2}$$$$-\frac{\mathrm{cosx}\{1+{(1-\mathrm{cosx})}^2\}-2\left\{\mathrm{sinx}\right\}\{1+{(1-\mathrm{cosx})}^2\}}{{\{1+{(1-\mathrm{cosx})}^2\}}^2}$$$$\Rightarrow {\lim }_{\mathrm{x}\to 0}{\mathrm{u}}^{\left(2\right)}\left(\mathrm{x}\right)=0-\frac{1}{1}=-1\mathrm{also}{\mathrm{v}}^{{}^{\prime }}\left(\mathrm{x}\right)=2\mathrm{x}\Rightarrow {\mathrm{v}}^{\left(2\right)}\left(\mathrm{x}\right)=2\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}{\mathrm{v}}^{\left(2\right)}\left(\mathrm{x}\right)=2\Rightarrow {\lim }_{\mathrm{x}\to 0}\frac{{\mathrm{u}}^{\left(2\right)}\left(\mathrm{x}\right)}{{\mathrm{v}}^{\left(2\right)}\left(\mathrm{x}\right)}=-\frac{1}{2}$$$$$$

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