We need to evaluate the given limit:

$\lim_{(x,y) \to (0,0)} y^2 \sin\left(\frac{1}{x}\right)$

Step-by-Step Solution:

1. Behavior of $\sin\left(\frac{1}{x}\right)$:

   • For $x \to 0$, $\frac{1}{x}$ oscillates between $-\infty$ and $+\infty$. Consequently, $\sin\left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ without settling to a single value.

2. Effect of $y^2$:

   • $y^2$ is non-negative and approaches $0$ as $y \to 0$.

3. Overall Product:

   • Since $y^2$ approaches $0$, the product $y^2 \sin\left(\frac{1}{x}\right)$ is squeezed between $-y^2$ and $+y^2$.
   • Therefore, irrespective of the oscillation of $\sin\left(\frac{1}{x}\right)$, the term $y^2 \sin\left(\frac{1}{x}\right)$ approaches $0$.

Conclusion:

The value of the limit is:

$\boxed{0}$

Let me know if you need further clarification or a deeper explanation!

Related Questions:

1. Why does the oscillation of $\sin(1/x)$ not affect the limit in this case?
2. How does the squeeze theorem apply to this problem?
3. Can we generalize this result for other functions of $y$?
4. What happens if the $y^2$ term were replaced with $y^n$ for $n > 2$?
5. How do polar coordinates simplify similar two-variable limits?

Tip: When working with multivariable limits, always analyze each term's contribution to the behavior of the overall expression.