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

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(B) To check for continuity at $x = 0$,we evaluate $\lim_{x 	o 0} f(x)$.Given $f(x) = x \sin \left( \frac{1}{x} ight) \sin \left( \frac{1}{x \sin \

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continuous but not differentiable at $x = 0$

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(B) To check for continuity at $x = 0$,we evaluate $\lim_{x 	o 0} f(x)$.Given $f(x) = x \sin \left( \frac{1}{x} ight) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} ight)} ight)$.Since $|\sin 	heta| \leq 1$,we have $|f(x)| = |x \sin \left( \frac{1}{x} ight) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} ight)} ight)| \leq |x \sin \left( \frac{1}{x} ight)| \leq |x|$.As $x 	o 0$,$|x| 	o 0$,so by the Squeeze Theorem,$\lim_{x 	o 0} f(x) = 0 = f(0)$. Thus,$f(x)$ is continuous at $x = 0$.To check for differentiability,we evaluate $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h \sin \left( \frac{1}{h} ight) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} ight)} ight)}{h} = \lim_{h 	o 0} \sin \left( \frac{1}{h} ight) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} ight)} ight)$.Let $g(h) = \sin \left( \frac{1}{h} ight) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} ight)} ight)$. As $h 	o 0$,$\sin \left( \frac{1}{h} ight)$ oscillates between $-1$ and $1$. The term $\frac{1}{h \sin \left( \frac{1}{h} ight)}$ also oscillates and takes arbitrarily large values,causing the second sine term to oscillate. Thus,the limit does not exist.Therefore,$f(x)$ is continuous but not differentiable at $x = 0$.

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

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