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Question Number 114259 by bobhans last updated on 18/Sep/20

$$\underset{x\to 0}{\lim }\frac{x\mathrm{arc}\sin \left(x^2\right)}{x\cos x-\sin x}?$$

Answered by bemath last updated on 18/Sep/20

$$by{L}^{\prime }Hopital$$$$\underset{x\to 0}{\lim }\frac{\mathrm{arc}\sin \left(x^2\right)+\frac{2x^2}{\sqrt{1-x^4}}}{\cos x-x\sin x-\cos x}=$$$$\underset{x\to 0}{\lim }\frac{\frac{\mathrm{arc}\sin \left(x^2\right)}{x^2}+\frac{2}{\sqrt{1-x^4}}}{-\frac{\sin x}{x}}=$$$$nowlet{L}_1=\underset{x\to 0}{\lim }\frac{\mathrm{arc}\sin \left(x^2\right)}{x^2}$$$$L_1=\underset{x\to 0}{\lim }\frac{\frac{2x}{\sqrt{1-x^4}}}{2x}=1$$$$soL=\underset{x\to 0}{\lim }\frac{\frac{\mathrm{arc}\sin \left(x^2\right)}{x^2}+\frac{2}{\sqrt{1-x^4}}}{-\frac{\sin x}{x}}$$$$L=\frac{1+2}{-1}=-3$$

Answered by Olaf last updated on 18/Sep/20

$$=\underset{x\to 0}{\lim }\frac{x\times x^2}{x(1-\frac{x^2}{2})-(x-\frac{x^3}{6})}$$$$=\underset{x\to 0}{\lim }\frac{x^3}{-\frac{x^3}{2}+\frac{x^3}{6}}=\frac{1}{-\frac{1}{2}+\frac{1}{6}}=-3$$$$$$

Commented by bemath last updated on 18/Sep/20

$$santuyy$$

Commented by malwaan last updated on 18/Sep/20

$$verysantuyygood$$

Commented by abdullahquwatan last updated on 18/Sep/20

$$mantep$$

Answered by mathmax by abdo last updated on 19/Sep/20

$$\mathrm{we}\mathrm{have}\mathrm{xarcsin}\left({\mathrm{x}}^2\right)\sim {\mathrm{x}}^3$$$$\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\Rightarrow \mathrm{xcosx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{2}\mathrm{and}\mathrm{sinx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{6}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^3}{\mathrm{x}-\frac{{\mathrm{x}}^3}{2}-\mathrm{x}+\frac{{\mathrm{x}}^3}{6}}=\frac{{\mathrm{x}}^3}{-\frac{1}{3}{\mathrm{x}}^3}=-3\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=-3$$

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