If f(x) = begin{cases} x sin frac{1}{x}, & x | vedclass

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(C) For the function $f(x)$ to be continuous at $x = 0$,the following condition must be satisfied:$f(0) = \lim_{x 	o 0} f(x)$Given $f(0) = k$,we need

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(C) For the function $f(x)$ to be continuous at $x = 0$,the following condition must be satisfied:$f(0) = \lim_{x 	o 0} f(x)$Given $f(0) = k$,we need to evaluate the limit:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left( x \sin \frac{1}{x} ight)$We know that for all $x 
e 0$,the sine function is bounded:$-1 \le \sin \frac{1}{x} \le 1$Multiplying by $x$ (where $x > 0$):$-x \le x \sin \frac{1}{x} \le x$Applying the Squeeze Theorem as $x 	o 0$:$\lim_{x 	o 0} (-x) = 0$ and $\lim_{x 	o 0} (x) = 0$Therefore,$\lim_{x 	o 0} x \sin \frac{1}{x} = 0$.Since $f(0) = k$,we have $k = 0$.

If $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

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