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To check the continuity of $f(x)$ at $x=0$, we need to verify if $\lim_{x \to 0} f(x) = f(0)$. Since $f(0) = 0$, we need to find $\lim_{x \to 0} x\sin\frac{1}{x}$.
We know that $-1 \leq \sin\frac{1}{x} \leq 1$ for all $x \neq 0$. Therefore, $-|x| \leq x\sin\frac{1}{x} \leq |x|$. As $x \to 0$, we have $-|x| \to 0$ and $|x| \to 0$. By the Squeeze Theorem, $\lim_{x \to 0} x\sin\frac{1}{x} = 0$. Since $\lim_{x \to 0} f(x) = 0 = f(0)$, the function $f(x)$ is continuous at $x=0$. Thus, Assertion (A) is true.
Reason (R) states that when $x \to 0$, $\sin\frac{1}{x}$ is a finite value between $-1$ and $1$. This statement is true because the sine function always oscillates between -1 and 1, regardless of how small $x$ becomes. However, the limit $\lim_{x \to 0} \sin\frac{1}{x}$ does not exist, but $\sin\frac{1}{x}$ remains bounded between -1 and 1.
The assertion is true, and the reason is also true. However, the reason correctly explains why the limit of $x\sin\frac{1}{x}$ as $x$ approaches 0 exists and is equal to 0, which makes the function continuous. Therefore, the reason is the correct explanation for the assertion.
Final Answer: Both A and R are true and R is the correct explanation of A.