91Ó°ÊÓ

Recall that a function \(f\) is continuous at a point \((x_0, y_0)\) if for every sequence \((x_n, y_n)\) that converges to \((x_0, y_0)\), the sequence \(f(x_n, y_n)\) converges to \(f(x_0, y_0)\). Essentially, it means the function has no gaps or jumps in its graph at the given point.

Observe the function given: $$f(x, y)=\sqrt{x^{2}+y^{2}}$$ This function represents the distance from the origin \((0,0)\) to the point \((x, y)\) in \(\mathbb{R}^{2}\). Since distance is a non-negative value and the function involves a square root, this function is always non-negative as well.

Consider any point \((x_0, y_0)\) in \(\mathbb{R}^{2}\). Let \((x_n, y_n)\) be a sequence of points that converges to \((x_0, y_0)\). We know that it will converge because the definition of continuity states that it is continuous if any sequence converging to \((x_0, y_0)\) also causes the function to converge to \(f(x_0, y_0)\). Now, we want to see if \(f(x_n, y_n)\) converges to \(f(x_0, y_0)\). We have $$f(x_n, y_n) = \sqrt{x_n^2 + y_n^2}$$ Since the sequence \((x_n, y_n)\) converges to \((x_0, y_0)\), we know that both \(x_n\) converges to \(x_0\) and \(y_n\) converges to \(y_0\). Thus, we can conclude that the sum \(x_n^2 + y_n^2\) converges to \(x_0^2 + y_0^2\) as well. Since the square root function is continuous as well, we know that the sequence \(f(x_n, y_n)\) converges to $$f(x_0, y_0) = \sqrt{x_0^2 + y_0^2}$$

Since we considered an arbitrary point \((x_0, y_0)\) in \(\mathbb{R}^{2}\) and showed that the function \(f(x,y) = \sqrt{x^2 + y^2}\) satisfies the definition of continuity at that point, we can conclude that the function is continuous at every point in \(\mathbb{R}^{2}\). So, the function $$f(x, y)=\sqrt{x^{2}+y^{2}}$$ is continuous at all points of \(\mathbb{R}^{2}\).