Let f(x) = begin{cases} x^2 sin left(frac{1}{ | vedclass

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(B) Continuity of $f(x)$ at $x=0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x^2 \sin \left(\frac{1}{x}ight)$. Since $|\sin(1/x)| \leq 1$,we have $|x^2

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$f$ is continuous but $f^{\prime}$ is not continuous

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(B) Continuity of $f(x)$ at $x=0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x^2 \sin \left(\frac{1}{x}ight)$. Since $|\sin(1/x)| \leq 1$,we have $|x^2 \sin(1/x)| \leq x^2$. By the Squeeze Theorem,$\lim_{x 	o 0} f(x) = 0 = f(0)$. Thus,$f(x)$ is continuous at $x=0$.Differentiability of $f(x)$ at $x=0$: $f^{\prime}(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h \sin(1/h) = 0$. Since the limit exists,$f(x)$ is differentiable at $x=0$ and $f^{\prime}(0) = 0$.Continuity of $f^{\prime}(x)$ at $x=0$: For $x 
eq 0$,$f^{\prime}(x) = \frac{d}{dx} [x^2 \sin(1/x)] = 2x \sin(1/x) + x^2 \cos(1/x) (-1/x^2) = 2x \sin(1/x) - \cos(1/x)$.As $x 	o 0$,$2x \sin(1/x) 	o 0$,but $\lim_{x 	o 0} \cos(1/x)$ does not exist due to oscillation. Therefore,$\lim_{x 	o 0} f^{\prime}(x)$ does not exist,so $f^{\prime}(x)$ is not continuous at $x=0$.

Let $f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{cases}$. Then at $x=0$:

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