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Solution

Here

\(f(x,y) = \frac{xy}{x^2 + y^2}\)

As (x, y) → (0, 0), we get f(x, y) → \(\frac{0}{0}\) form

So, set y = mx for x ≠ 0 where m is some constant value then

\(f(x,y) = \frac{mx^2}{x^2 + (mx)^2}\)

\(= \frac{mx^2}{x^2(1 + m^2)}\)

\(= \frac{m}{1 + m^2}\)

Since y is variable, so m has no fixed value. That means f(x, y) may not have a fixed value as (x, y) → (0, 0).

This means, f(x, y) is not necessarily zero at (x, y) ≠ (0, 0). That is f(x, y) is not continuous at (x, y) = (0, 0).

But \(f(x,y) = \frac{xy}{x^2 + y^2}\) for f(x, y) ≠ (0, 0) that is defined single form function except at (0, 0).

Since xy and \(x^2 + y^2\) both are polynomial functions which are continuous, so f(x, y) is continuous except at \(x^2 + y^2 = 0\). This means f(x, y) is not continuous at (0, 0).

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