To show that the function $f(x, y) = \frac{2xy}{x^2 + y^2}$ is continuous, we need to demonstrate that it is continuous at every point in its domain, especially at points where the denominator might be zero, i.e., at $(x, y) = (0, 0)$.

Step-by-Step Analysis:

1. Domain of the Function:
   The function is defined as: $f(x, y) = \frac{2xy}{x^2 + y^2}$ The denominator $x^2 + y^2$ is zero only at $(x, y) = (0, 0)$. Therefore, $f(x, y)$ is not defined at $(0, 0)$. To extend the function to include $(0, 0)$, we will define: $f(0, 0) = 0.$

2. Continuity Away from the Origin:
   For any point $(x, y) \neq (0, 0)$, the denominator $x^2 + y^2$ is positive, so the function is the ratio of two continuous functions (a polynomial in the numerator and a nonzero quadratic form in the denominator). Thus, $f(x, y)$ is continuous for $(x, y) \neq (0, 0)$.

3. Continuity at the Origin:
   To prove that the function is continuous at the origin, we need to check the limit: $\lim_{(x, y) \to (0, 0)} f(x, y) = 0$ If this limit exists and equals $f(0, 0)$, which we have defined as 0, then the function will be continuous at $(0, 0)$.

   Let's calculate the limit in polar coordinates, where $x = r\cos\theta$ and $y = r\sin\theta$, and $r = \sqrt{x^2 + y^2}$: $f(x, y) = \frac{2xy}{x^2 + y^2} = \frac{2r^2\cos\theta\sin\theta}{r^2} = 2\cos\theta\sin\theta = \sin(2\theta)$ As $r \to 0$, the value of the function does not depend on $r$, and $\sin(2\theta)$ is bounded. Therefore, the limit is: $\lim_{r \to 0} \sin(2\theta) = 0$ since $r \to 0$ implies that $x$ and $y$ approach 0.

4. Conclusion:
   The function $f(x, y) = \frac{2xy}{x^2 + y^2}$ is continuous at all points, including at $(0, 0)$, where it approaches the value $0$.

Thus, the function is continuous.

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Would you like more details on the steps or have any other questions?

Here are some related questions:

1. What is the definition of continuity for a multivariable function?
2. How do we convert Cartesian coordinates to polar coordinates?
3. Why do we check the limit in polar coordinates?
4. How does the function behave along different paths approaching $(0, 0)$?
5. What is a removable discontinuity, and does it apply here?

Tip: When checking the continuity of a function at the origin, switching to polar coordinates often simplifies the analysis by reducing the problem to a radial limit.