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We need to find the limit of $f(x)$ as $x$ approaches 0 from the left side. $$LHL = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x \sin(\frac{1}{x})$$ Since $-1 \le \sin(\frac{1}{x}) \le 1$, we have $-|x| \le x \sin(\frac{1}{x}) \le |x|$. As $x \to 0^-$, $|x| \to 0$. Therefore, by the Squeeze Theorem, $$LHL = \lim_{x \to 0^-} x \sin(\frac{1}{x}) = 0$$
We need to find the limit of $f(x)$ as $x$ approaches 0 from the right side. $$RHL = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x \sin(\frac{1}{x})$$ Since $-1 \le \sin(\frac{1}{x}) \le 1$, we have $-|x| \le x \sin(\frac{1}{x}) \le |x|$. As $x \to 0^+$, $|x| \to 0$. Therefore, by the Squeeze Theorem, $$RHL = \lim_{x \to 0^+} x \sin(\frac{1}{x}) = 0$$
The function is defined as $f(0) = 0$.
We have $LHL = 0$, $RHL = 0$, and $f(0) = 0$. Since $LHL = RHL = f(0)$, the function is continuous at $x=0$.
Final Answer: The function is continuous at x=0.
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