Let f, g: R rightarrow R be functions defined | vedclass

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(B) For $f(x)$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x \sin(1/x) = 0$,which equals $f(0)$,so $f(x)$ is continuous at $x = 0$. For differentiability a

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Both $(i)$ and $(ii)$ are true

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(B) For $f(x)$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x \sin(1/x) = 0$,which equals $f(0)$,so $f(x)$ is continuous at $x = 0$. For differentiability at $x = 0$,$f'(0) = \lim_{h 	o 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h 	o 0} \sin(1/h)$,which does not exist. Thus,$(i)$ is true. For $g(x) = x f(x) = x^2 \sin(1/x)$ for $x 
eq 0$ and $g(0) = 0$. $g'(0) = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h \sin(1/h) = 0$. So $g(x)$ is differentiable at $x = 0$. For $x 
eq 0$,$g'(x) = 2x \sin(1/x) - \cos(1/x)$. As $x 	o 0$,$\lim_{x 	o 0} g'(x) = \lim_{x 	o 0} (2x \sin(1/x) - \cos(1/x))$,which does not exist because $\cos(1/x)$ oscillates. Since $\lim_{x 	o 0} g'(x) 
eq g'(0)$,$g'(x)$ is not continuous at $x = 0$. Thus,$(ii)$ is true.

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