Explanation

(a)
$\lim_{x \rightarrow 0}x\log{\sin x} = \lim_{x \rightarrow 0}\frac{\log{\sin x}}{1/x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{0}{0}\ form \right)$
$= \lim_{x \rightarrow 0}\ \frac{\frac{1}{\sin x}\cos x}{\frac{- 1}{x^{2}}} = \lim_{x \rightarrow 0} - \frac{x^{2}}{\tan x}$ [by L’Hospital’s rule]
$= \lim_{x \rightarrow 0}\ \frac{- 2x}{\sec^{2}x}$ [by L’Hospital’s rule]
$= 0$