If the function f(x) defined by f(x) = begin{ | vedclass

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(A) Given the function $f(x) = \begin{cases} x \sin \frac{1}{x}, & x 
eq 0 \ k, & x = 0 \end{cases}$.Since $f(x)$ is continuous at $x = 0$,the condi

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(A) Given the function $f(x) = \begin{cases} x \sin \frac{1}{x}, & x 
eq 0 \ k, & x = 0 \end{cases}$.Since $f(x)$ is continuous at $x = 0$,the condition $\lim_{x 	o 0} f(x) = f(0)$ must hold.We calculate the limit: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x \sin \frac{1}{x}$.We know that for all $x 
eq 0$,$-1 \leq \sin \frac{1}{x} \leq 1$.Multiplying by $x$,we get $-|x| \leq x \sin \frac{1}{x} \leq |x|$.By the Squeeze Theorem,as $x 	o 0$,both $-|x|$ and $|x|$ approach $0$.Therefore,$\lim_{x 	o 0} x \sin \frac{1}{x} = 0$.Since $f(0) = k$,we have $k = 0$.

If the function $f(x)$ defined by $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k = . . . . . .$

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