bemath-lim-x-0-1-x-sin-1-x-1-x-2-

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

\sqrt{bemath}

\lim_{x \to 0} (\frac{1}{x \sin^{- 1} (x)} - \frac{1}{x^{2}}) = ?

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

J \overset{- - - - - - - -}{S} ★

l e t \sin^{- 1} (x) = p \Rightarrow x = \sin p

\lim_{p \to 0} (\frac{1}{p \sin p} - \frac{1}{\sin^{2} p}) =

\lim_{p \to 0} (\frac{\sin p - p}{p \sin^{2} p}) =

\lim_{p \to 0} (\frac{\cos p - 1}{3 p^{2}}) = \lim_{p \to 0} (\frac{- 2 \sin^{2} (\frac{p}{2})}{3 p^{2}})

= - \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

\lim_{x \to 0} (\frac{1}{x (x + \frac{x^{3}}{6})} - \frac{1}{x . x}) = \frac{1}{x} (\frac{x - x - \frac{x^{3}}{6}}{x (x + \frac{x^{3}}{6})}) = \frac{- \frac{x}{6}}{x + \frac{x^{3}}{6}} = - \frac{1}{6}