Question-178178

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

Answered by Frix last updated on 14/Oct/22

\lim_{x \to 0} \frac{x - \sin^{- 1} x}{x^{2} \sin^{- 1} x} =_{[t = \sin^{- 1} x]}

= - \lim_{t \to 0} \frac{t - \sin t}{t \sin^{2} t} =_{[3 times l^{'} H \hat{o p i t a l}]}

= - \lim_{t \to 0} \frac{\frac{d^{3} [t - \sin t]}{{d t}^{3}}}{\frac{d^{3} [t \sin^{2} t]}{{d t}^{3}}} =

= - \lim_{t \to 0} \frac{\cos t}{{12 \cos}^{2} t - 8 t \cos t \sin t - 6} =

= - \frac{1}{6}

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