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Question Number 113747 by deleteduser12 last updated on 15/Sep/20

$$\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x}{x}$$

Answered by bemath last updated on 15/Sep/20

$$\underset{x\to 0}{\lim }\frac{d\left({\sin }^{-1}x\right)}{dx}=\underset{x\to 0}{\lim }\frac{\frac{1}{\sqrt{1-x^2}}}{1}=1$$

Answered by Olaf last updated on 15/Sep/20

$$x=\sin u$$$$\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x}{x}=\underset{u\to 0}{\lim }\frac{u}{\sin u}=1$$$$\left(\mathrm{usual}\mathrm{limit}\right)$$

Commented by Ar Brandon last updated on 15/Sep/20

Salut monsieur. Heureux de vous revoir��

Answered by Dwaipayan Shikari last updated on 15/Sep/20

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

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