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(B) For $f(x)$ at $x=0$:$LHL = \lim_{h 	o 0^-} (-h) \sin(-1/h) = \lim_{h 	o 0^-} h \sin(1/h) = 0$.$RHL = \lim_{h 	o 0^+} h \sin(1/h) = 0$.Since $f(

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Both statement $I$ and $II$ are true.

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(B) For $f(x)$ at $x=0$:$LHL = \lim_{h 	o 0^-} (-h) \sin(-1/h) = \lim_{h 	o 0^-} h \sin(1/h) = 0$.$RHL = \lim_{h 	o 0^+} h \sin(1/h) = 0$.Since $f(0) = 0$,$LHL = RHL = f(0)$,so $f$ is continuous at $x=0$. Statement $I$ is true.For $g(x) = x f(x) = \begin{cases} x^2 \sin(1/x), & x 
e 0 \ 0, & x = 0 \end{cases}$.To check differentiability at $x=0$,we find $g'(0) = \lim_{h 	o 0} \frac{g(h) - g(0)}{h} = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h \sin(1/h) = 0$.Since the limit exists and is finite,$g$ is differentiable at $x=0$. Statement $II$ is true.

Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.
Statement $I$: $f$ is a continuous function at $x = 0$.
Statement $II$: $g$ is a differentiable function at $x = 0$.

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Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.Statement $I$: $f$ is a continuous function at $x = 0$.Statement $II$: $g$ is a differentiable function at $x = 0$.