Problem 34 - Calculus Early Transcendentals

Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\frac{x y}{x^{2} y^{2}+1}$$

Step-by-Step Solution

Answer

Answer: Yes, the function is continuous for all points in \(\mathbb{R}^{2}\).

Step 1: Analyze the denominator

The denominator of the function is given by \(x^{2} y^{2} + 1\). Notice that for any point \((x, y)\) in \(\mathbb{R}^{2}\), the denominator will always be different from zero, because the sum of squares is always non-negative and we are adding 1 to it. Thus, \(x^{2} y^{2}+1 > 0\) for all \((x, y) \in \mathbb{R}^{2}\).

Step 2: Use the properties of continuous functions

The functions \(x\) and \(y\) are continuous on \(\mathbb{R}^{2}\), and the multiplication of continuous functions is also continuous. So the function \(x^2 y^2\) is continuous for each \((x, y)\) in \(\mathbb{R}^2\). Furthermore, the function \(x^{2} y^{2}+1\) is continuous since it is a sum of continuous functions \(x^2 y^2\) and a constant function 1.

Step 3: Find the limit as \((x, y)\) approaches any point \((a, b)\)

Given that both numerator and denominator are continuous, we have: \(\lim_{(x,y) \to (a,b)} f(x, y) = \lim_{(x,y) \to (a,b)} \frac{x y}{x^{2} y^{2}+1} = \frac{\lim_{(x,y) \to (a,b)} (x y)}{\lim_{(x,y) \to (a,b)} (x^{2} y^{2}+1)} = \frac{ab}{(a^2 b^2 + 1)} = f(a, b)\). Since \(\lim_{(x,y) \to (a,b)} f(x, y) = f(a, b)\), the function is continuous at each point \((a, b)\) in \(\mathbb{R}^2\). In conclusion, the function \(f(x,y) = \frac{x y}{x^{2} y^{2}+1}\) is continuous at all points of \(\mathbb{R}^{2}\).