Explanation

(a)
$\lim_{x \rightarrow 0}\frac{\tan{x - \sin x}}{x^{3}} = \lim_{x \rightarrow 0}\frac{\sec^{2}x - \cos x}{3x^{2}}$
[using L’Hospital’s rule]
$= \lim_{x \rightarrow 0}\frac{{2sec}^{2}x\tan x + \sin x}{6x}$
[using L’Hospital’s rule]
$= \lim_{x \rightarrow 0}\frac{\left\lbrack {2(\sec^{2}}x\sec^{2}{x + 2\sec{x \times \sec{x\tan x}}}\tan{x) +}\cos x \right\rbrack}{6}$
[by L’Hospital’s rule]
$= \frac{2\lbrack 1.1 + 2(0) + 1\rbrack}{6} = \frac{1}{2}$