91Ó°ÊÓ

For all points \((x, y) \neq (0, 0)\), we know that the function is defined as \(f(x, y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2}\). This is a rational function with a continuous numerator and continuous denominator in the domain \(\mathbb{R}^2\) except at \((0,0)\). This means that the function will be continuous at all points \((x, y) \neq (0, 0)\) in \(\mathbb{R}^2\).

Now, we need to check the continuity of the function at the point \((0, 0)\). The function is defined as \(f(0, 0) = 1\). To check the continuity, we will find the limit of the function as \((x, y)\) approaches \((0, 0)\). $$\lim_{(x, y) \to (0, 0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2}$$ We use the polar coordinates transformation, where \(x = r\cos\theta\) and \(y = r\sin\theta\). Since \(x^2 + y^2 = r^2\), we rewrite the limit in polar coordinates: $$\lim_{r \to 0} \frac{\sin(r^2)}{r^2}$$ Now, using the standard limit \(\lim_{z \to 0} \frac{\sin z}{z} = 1\), we can replace \(z = r^2\) and obtain: $$\lim_{r \to 0} \frac{\sin(r^2)}{r^2} = 1$$ Since the limit of the function exists and is equal to \(1\), the same value as \(f(0, 0)\), the function is continuous at \((0, 0)\).

The function \(f(x, y)\) is continuous at all points in \(\mathbb{R}^2\), including \((0, 0)\).