$\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)$

Step 1 ：The limit of a product is the product of the limits, provided that the limits exist. In this case, we have the product of \(x\) and \(\sin\left(\frac{1}{x}\right)\) as \(x\) approaches 0. The limit of \(x\) as \(x\) approaches 0 is 0. However, the limit of \(\sin\left(\frac{1}{x}\right)\) as \(x\) approaches 0 does not exist because the function oscillates between -1 and 1 infinitely often as \(x\) approaches 0. Therefore, we cannot directly apply the limit of a product rule.

Step 2 ：We can use the squeeze theorem (also known as the sandwich theorem) to solve this problem. The squeeze theorem states that if we have three functions, \(f(x)\), \(g(x)\), and \(h(x)\), and if \(f(x) \leq g(x) \leq h(x)\) for all \(x\) in an interval around a point \(a\), except possibly at \(a\) itself, and if \(\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} h(x) = L\), then \(\lim_{x \rightarrow a} g(x) = L\).

Step 3 ：In this case, we can use the fact that \(-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\) for all \(x\) to say that \(-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|\). As \(x\) approaches 0, both \(-|x|\) and \(|x|\) approach 0, so by the squeeze theorem, \(x \sin\left(\frac{1}{x}\right)\) also approaches 0.

Step 4 ：Final Answer: \(\boxed{0}\)