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Question Number 111326 by bemath last updated on 03/Sep/20

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

Answered by john santu last updated on 03/Sep/20

$$J\stackrel{--------}{S}★$$$$let{\sin }^{-1}\left(x\right)=p\Rightarrow x=\sin p$$$$\underset{p\to 0}{\lim }(\frac{1}{p\sin p}-\frac{1}{\sin {}^{2}p})=$$$$\underset{p\to 0}{\lim }\left(\frac{\sin p-p}{p\sin {}^{2}p}\right)=$$$$\underset{p\to 0}{\lim }\left(\frac{\cos p-1}{3p^2}\right)=\underset{p\to 0}{\lim }\left(\frac{-2\sin {}^2\left(\frac{p}{2}\right)}{3p^2}\right)$$$$=-\frac{2}{3}\times \frac{1}{4}=-\frac{1}{6}$$

Commented by bemath last updated on 03/Sep/20

Answered by Dwaipayan Shikari last updated on 03/Sep/20

$$\underset{x\to 0}{\lim }(\frac{1}{x(x+\frac{x^3}{6})}-\frac{1}{x.x})=\frac{1}{x}\left(\frac{x-x-\frac{x^3}{6}}{x(x+\frac{x^3}{6})}\right)=\frac{-\frac{x}{6}}{x+\frac{x^3}{6}}=-\frac{1}{6}$$

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