$\underset{x\to 0}{lim}x\sin \left(\frac{1}{x}\right)$

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 $f(x)\le g(x)\le h(x)$ for all $x$ in an interval around a point $a$, except possibly at $a$ itself, and if $\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}h(x)=L$, then $\underset{x\to a}{lim}g(x)=L$.

Step 3 ：In this case, we can use the fact that $-1\le \sin \left(\frac{1}{x}\right)\le 1$ for all $x$ to say that $-|x|\le x\sin \left(\frac{1}{x}\right)\le |x|$. 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: $\boxed{{\displaystyle 0}}$