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Question Number 178178 by cortano1 last updated on 13/Oct/22

Answered by Frix last updated on 14/Oct/22

$$\underset{x\to 0}{\lim }\frac{x-{\sin }^{-1}x}{x^2{\sin }^{-1}x}{=}_{[t={\sin }^{-1}x]}$$$$=-\underset{t\to 0}{\lim }\frac{t-\sin t}{t{\sin }^{2}t}{=}_{\left[3\mathrm{times}{\mathrm{l}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}\right]}$$$$=-\underset{t\to 0}{\lim }\frac{\frac{d^3[t-\sin t]}{{dt}^3}}{\frac{d^3\left[t{\sin }^{2}t\right]}{{dt}^3}}=$$$$=-\underset{t\to 0}{\lim }\frac{\cos t}{{12\cos }^{2}t-8t\cos t\sin t-6}=$$$$=-\frac{1}{6}$$

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