Answer
a) Cannot apply the L'Hospital's rule
b) $0$
Work Step by Step
a) Our aim is to evaluate the limit for $\lim\limits_{x \to 0} \dfrac{x^2 \sin (1/x)}{\sin x}$
But $\lim\limits_{x \to 0} \dfrac{x^2 \sin (1/x)}{\sin x}$ does not show any indeterminate form . So, we cannot apply the L'Hospital's rule .
b) Our aim is to evaluate the limit for $\lim\limits_{x \to 0} \dfrac{x^2 \sin (1/x)}{\sin x}$ by the inspection method.
$\lim\limits_{x \to 0} \dfrac{x^2 \sin (1/x)}{\sin x}=\lim\limits_{x \to 0} \dfrac{x}{\sin x } \times \lim\limits_{x \to 0} x \times \lim\limits_{x \to 0} \sin \dfrac{1}{x}\\=(1) \times \lim\limits_{x \to 0} \dfrac{\sin (1/x)}{1/x}\\=1 \times 0\\=0$
