# If $$f(x,y) = \frac{xy}{x^2 + y^2}$$, does f(x, y) exist, as (x, y) → (0, 0)?

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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).