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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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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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