Let f(x) = begin{cases} (3 - sin(1/x))|x|, & | vedclass

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(B) First,we check the continuity of $f(x)$ at $x = 0$. $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} (3 - \sin(1/x))|x|$. Since $|\sin(1/x)| \le 1$,we have $

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(B) First,we check the continuity of $f(x)$ at $x = 0$. $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} (3 - \sin(1/x))|x|$. Since $|\sin(1/x)| \le 1$,we have $2|x| \le (3 - \sin(1/x))|x| \le 4|x|$. By the Sandwich Theorem,$\lim_{x 	o 0} f(x) = 0 = f(0)$. Thus,$f$ is continuous at $x = 0$. Next,we examine the values of $f(x)$ near $x = 0$. For any $x 
e 0$,$|x| > 0$ and $(3 - \sin(1/x)) \ge 3 - 1 = 2 > 0$. Therefore,$f(x) = (3 - \sin(1/x))|x| > 0$ for all $x 
e 0$. Since $f(0) = 0$ and $f(x) > 0$ for all $x 
e 0$ in the neighborhood of $0$,$f(x) \ge f(0)$ for all $x$ in the neighborhood. Hence,$f$ has a local minimum at $x = 0$.

Let $f(x) = \begin{cases} (3 - \sin(1/x))|x|, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Then at $x = 0$,$f$ has a

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