The given expression is:

$\lim_{(x, y) \to (0, 0)} \frac{x^2 \sin^2(y)}{x^2 + 3y^2}$

To solve this limit using polar coordinates, we substitute:

• $x = r \cos(\theta)$
• $y = r \sin(\theta)$

where $r$ is the radial distance from the origin, and $\theta$ is the angle. In polar coordinates, as $(x, y) \to (0, 0)$, this corresponds to $r \to 0$.

Let's rewrite the given limit in polar coordinates:

• $x^2 = r^2 \cos^2(\theta)$
• $y^2 = r^2 \sin^2(\theta)$
• $\sin^2(y) = \sin^2(r \sin(\theta))$

Thus, the expression becomes:

$\frac{x^2 \sin^2(y)}{x^2 + 3y^2} = \frac{r^2 \cos^2(\theta) \sin^2(r \sin(\theta))}{r^2 \cos^2(\theta) + 3r^2 \sin^2(\theta)}$

Now, simplify:

$\frac{r^2 \cos^2(\theta) \sin^2(r \sin(\theta))}{r^2(\cos^2(\theta) + 3 \sin^2(\theta))}$

Factor out $r^2$ from both the numerator and denominator:

$= \frac{\cos^2(\theta) \sin^2(r \sin(\theta))}{\cos^2(\theta) + 3 \sin^2(\theta)}$

As $r \to 0$, $\sin^2(r \sin(\theta))$ approaches $0$, since $\sin(r \sin(\theta)) \approx r \sin(\theta)$ when $r$ is small. Hence:

$\sin^2(r \sin(\theta)) \to 0$

Therefore, the entire expression approaches:

$\frac{0}{\cos^2(\theta) + 3 \sin^2(\theta)} = 0$

Thus, the limit is:

$\boxed{0}$

Would you like more details on this calculation or to ask any follow-up questions?

Here are 5 related questions you might find useful:

1. How do polar coordinates help in simplifying limits?
2. What is the significance of the small-angle approximation for sine in this solution?
3. Can this method be applied to any multivariable limit problem?
4. How would you compute limits that approach non-origin points using polar coordinates?
5. How can L'Hopital's rule be used in multivariable limit problems?

Tip: Polar coordinates are particularly useful in problems involving circular symmetry around the origin, as they can simplify expressions involving powers of $x$ and $y$.