The function f(x) = x^2 sin frac{1}{x} for x | vedclass

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(D) To check continuity at $x = 0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x^2 \sin \frac{1}{x}$. Since $-1 \le \sin \frac{1}{x} \le 1$,we have $-x^2 \

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Is continuous and differentiable

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(D) To check continuity at $x = 0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x^2 \sin \frac{1}{x}$. Since $-1 \le \sin \frac{1}{x} \le 1$,we have $-x^2 \le x^2 \sin \frac{1}{x} \le x^2$. By the squeeze theorem,$\lim_{x 	o 0} f(x) = 0 = f(0)$. Thus,$f(x)$ is continuous at $x = 0$. To check differentiability at $x = 0$: $f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h \sin \frac{1}{h}$. Since $\lim_{h 	o 0} h = 0$ and $\sin(1/h)$ is bounded,the limit is $0$. Thus,$f'(0) = 0$,so the function is differentiable at $x = 0$.

The function $f(x) = x^2 \sin \frac{1}{x}$ for $x \ne 0$ and $f(0) = 0$ at $x = 0$ is:

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