91Ó°ÊÓ

A function \(f(x, y)\) is continuous at a point \((a, b)\) if the limit of the function exists at that point and the value of the function at that point is equal to the limit, i.e., \(\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b)\). We need to determine whether this property holds true for the given function \(p(x, y)\) at all points in \(\mathbb{R}^2\).

First, we need to find out where the function is not defined as it cannot be continuous in those points. The function \(p(x, y)\) would not be defined if the denominator is zero. So, we need to check when \(x^4 + y^2 = 0\). However, the equation \(x^4 + y^2 = 0\) has no solution because \(x^4 \geq 0\) and \(y^2 \geq 0\), which implies that \(x^4 + y^2 \geq 0\). Therefore, the only time the denominator can be zero is if \(x^4 = y^2 = 0\), which is only possible if \(x = y = 0\). Hence, \(p(x, y)\) is undefined only at \((0, 0)\).

Since the function is defined everywhere except \((0, 0)\), we will now look at the continuity for points other than \((0, 0)\). For \(x^4 + y^2 \neq 0\), by using basic limit properties and the fact that the limit of a rational function exists and equals the function value at all points where it is defined, we can find that \(\lim_{(x, y) \to (a, b)} p(x, y) = \frac{4a^2b^2}{a^4 + b^2} = p(a, b)\) for \((a, b) \neq (0, 0)\), so \(p(x, y)\) is continuous at all points \((a, b) \neq (0, 0)\).

Finally, we need to check for continuity at the point \((0, 0)\). Let's determine the limit of the function as \((x, y) \to (0, 0)\). To do this, we will look at the limit of \(p(x, y)\) along different paths approaching \((0, 0)\). Let's first consider the path along the \(x\)-axis (\(y = 0\)): $$\lim_{x \to 0} p(x, 0) = \lim_{x \to 0} \frac{4 x^{2} (0)^{2}}{x^{4}+(0)^{2}} = 0$$ Now let's consider the path along the \(y\)-axis (\(x = 0\)): $$\lim_{y \to 0} p(0, y) = \lim_{y \to 0} \frac{4 (0)^{2} y^{2}}{(0)^{4}+y^{2}} = 0$$ We see that the limits along the \(x\)-axis and \(y\)-axis are both 0. Let's now consider the limit along the line \(y = x\): $$\lim_{x \to 0} p(x, x) = \lim_{x \to 0} \frac{4 x^{2} x^{2}}{x^{4} + x^{2}} = \lim_{x \to 0} \frac{4x^4}{x^4 + x^2}$$ divide numerator and denominator by x^2 (which we can do since we're finding the limit as \(x \to 0\)) and get: $$\lim_{x \to 0} \frac{4x^2}{x^2 + 1} = 0$$ Since the limit of the function is the same along all three paths, we can conclude that the limit of the function as \((x, y) \to (0, 0)\) exists and is 0, i.e., \(\lim_{(x, y) \to (0, 0)} p(x, y) = 0\). However, \(p(0,0)\) is undefined, so the function cannot be continuous at \((0, 0)\).

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