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The given limit problem is:

\[
\lim_{(x, y) \to (0, 0)} \frac{x^3 y - 2y}{x^2 + xy + y^2}
\]

### Step 1: Analyzing the Limit
We need to evaluate the limit as \( (x, y) \) approaches \( (0, 0) \). This is a multivariable limit, so the approach to zero can vary based on the path we choose. To verify whether the limit exists, we can evaluate the limit along different paths.

### Step 2: Path 1: Along \( y = 0 \)
When \( y = 0 \), the expression becomes:

\[
\frac{x^3(0) - 2(0)}{x^2 + x(0) + (0)^2} = \frac{0}{x^2} = 0
\]

So, along \( y = 0 \), the limit is 0.

### Step 3: Path 2: Along \( x = 0 \)
When \( x = 0 \), the expression becomes:

\[
\frac{0^3(y) - 2y}{0^2 + 0(y) + y^2} = \frac{-2y}{y^2} = \frac{-2}{y}
\]

As \( y \to 0 \), \( \frac{-2}{y} \) does not approach a finite value; instead, it goes to infinity or negative infinity depending on the direction from which \( y \) approaches 0. Therefore, the limit does not exist along this path.

### Conclusion:
Since the limit behaves differently along different paths, the overall limit does not exist.

### Would you like more details or have any questions about the steps?

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Here are 5 related questions to further your understanding:
1. How do you determine if a multivariable limit exists?
2. Why is evaluating limits along different paths important for multivariable limits?
3. What does it mean if a limit approaches different values along different paths?
4. Can the polar coordinate substitution method help for other multivariable limit problems?
5. What is the definition of continuity for a function of multiple variables?

**Tip:** When dealing with multivariable limits, always try evaluating the function along different paths to check if the limit is path-dependent.

Evaluate the limit: lim (x, y) → (0, 0) (x^3y - 2y) / (x^2 + xy + y^2)

Learn how to evaluate the multivariable limit lim (x, y) → (0, 0) (x^3y - 2y) / (x^2 + xy + y^2) with a step-by-step solution. This problem explores path-dependence in limits and is suitable for undergraduate-level calculus students.

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