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Chapter 12: Problem 38

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

Short Answer

Expert verified
Answer: The function is continuous at all points (x, y) 鈭� 鈩澛� except for the points where the denominator is equal to zero, which are x=0, y=-1, and y=1.

Step by step solution

01

Determine the points of potential discontinuity

To find the points of potential discontinuity, we need to find the points where the denominator is equal to zero. Let's consider the denominator part, which is: $$x\left(y^{2}-1\right) = x(y+1)(y-1)$$ The denominator becomes zero when \(x=0\), \(y=-1\), or \(y=1\). So the problematic points are: 1. \(x=0\) 2. \(y=-1\) 3. \(y=1\)
02

Check for continuity at other points

Now, we need to verify if \(f(x, y)\) is continuous at all other points \((x, y) \in \mathbb{R}^2\), excluding the ones found in Step 1. To verify this, we will use the limit definition of continuity: $$\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b)$$ In our case, \((a, b) \ne (0, -1, 1)\), which means we are not considering the points where the denominator is equal to zero. The function \(f(x, y)\) is a rational function, i.e., a quotient of two polynomials. Rational functions are continuous everywhere except where the denominator is equal to zero. Hence, \(f(x, y)\) is continuous for all other points in \(\mathbb{R}^2\) except for the problematic points found in Step 1.
03

Conclusion

The given function \(f(x, y)\) is continuous at all points \((x, y) \in \mathbb{R}^2\) except for the points where the denominator is equal to zero, which are \(x=0\), \(y=-1\), and \(y=1\).

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