https://file.aiquickdraw.com/mathbot/file/1728975809876uxpkhf7i.jpg

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 \(y\).

Calculate the limit: lim_{(x, y) -> (0, 0)} (x^2 sin^2(y)) / (x^2 + 3y^2) using polar coordinates.

Learn how to solve a multivariable limit lim_{(x, y) -> (0, 0)} (x^2 sin^2(y)) / (x^2 + 3y^2) using polar coordinates. This problem focuses on polar coordinate substitution and simplifies the expression step-by-step. Suitable for students in advanced calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation