Explanation

(c)
$\lim_{x \rightarrow 0}\frac{\sin^{- 1}{x - x}}{x^{3}\cos x} = \lim_{x \rightarrow 0}\frac{\frac{1}{\sqrt{1 - x^{2}}} - 1}{x^{3}\left( - \sin x \right) + 3x^{2}\cos x}$
[using L’Hospital’s rule]
$= \lim_{x \rightarrow 0}\frac{1 - \sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}}.x^{2}( - x\sin{x + 3\cos x})} \times \frac{1 + \sqrt{1 - x^{2}}}{1 + \sqrt{1 - x^{2}}}$
$= \lim_{x \rightarrow 0}\frac{1}{\left\lbrack \begin{array}{r}
\sqrt{1 - x^{2}}\left( 1 + \sqrt{1 - x^{2}} \right) \\
( - x\sin{x + 3\cos x})
\end{array} \right\rbrack}$
$= \frac{1}{1(1 + 1)(3)} = \frac{1}{6}$