The problem asks to determine the largest set S on which the given functions f are continuous. We need to consider the domain of each function and identify points where the function is not defined or might have discontinuities.

1. Problem Description

The problem asks to determine the largest set S on which the given functions f are continuous. We need to consider the domain of each function and identify points where the function is not defined or might have discontinuities.

2. Solution Steps

1

7. $f(x, y) = \frac{x^2 + xy - 5}{x^2 + y^2 + 1}$

The denominator $x^2 + y^2 + 1$ is always positive and never zero. The numerator is a polynomial. Thus, the function is continuous everywhere.

S = $R^2$

1

8. $f(x, y) = \ln(1 + x^2 + y^2)$

The natural logarithm is continuous where its argument is positive. Here, the argument is $1 + x^2 + y^2$, which is always greater than or equal to

1. Therefore, the function is continuous everywhere.

S = $R^2$

1

9. $f(x, y) = \ln(1 - x^2 - y^2)$

For the logarithm to be defined and continuous, we require $1 - x^2 - y^2 > 0$, which means $x^2 + y^2 < 1$. This represents the open disk centered at the origin with radius

1. S = $\{(x, y) \in R^2 | x^2 + y^2 < 1\}$

2

0. $f(x, y) = \frac{1}{\sqrt{1 + x + y}}$

For the function to be defined and continuous, we need the expression inside the square root to be positive, i.e., $1 + x + y > 0$, which means $y > -x - 1$.

S = $\{(x, y) \in R^2 | y > -x - 1\}$

2

1. $f(x, y) = \frac{x^2 + 3xy + y^2}{y - x^2}$

The function is continuous as long as the denominator is not zero, i.e., $y - x^2 \neq 0$, or $y \neq x^2$.

S = $\{(x, y) \in R^2 | y \neq x^2\}$

2

2. $f(x, y) = \begin{cases} \frac{\sin(xy)}{xy}, & \text{if } xy \neq 0 \\ 1, & \text{if } xy = 0 \end{cases}$

If $xy \neq 0$, then $f(x, y) = \frac{\sin(xy)}{xy}$. We know that $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$. Therefore, even as $xy$ approaches 0, the limit of $\frac{\sin(xy)}{xy}$ is 1, which matches the definition of the function when $xy = 0$. Therefore, the function is continuous everywhere.

S = $R^2$

2

3. $f(x, y) = \sqrt{x - y + 1}$

For the square root to be defined and continuous, we require $x - y + 1 \geq 0$, which means $y \leq x + 1$.

S = $\{(x, y) \in R^2 | y \leq x + 1\}$

2

4. $f(x, y) = (4 - x^2 - y^2)^{-1/2} = \frac{1}{\sqrt{4 - x^2 - y^2}}$

For the function to be defined and continuous, we need the expression inside the square root to be positive, i.e., $4 - x^2 - y^2 > 0$, which means $x^2 + y^2 < 4$. This represents the open disk centered at the origin with radius

2. S = $\{(x, y) \in R^2 | x^2 + y^2 < 4\}$

2

5. $f(x, y, z) = \frac{1 + x^2}{x^2 + y^2 + z^2}$

The function is continuous as long as the denominator is not zero, i.e., $x^2 + y^2 + z^2 \neq 0$, which means $(x, y, z) \neq (0, 0, 0)$.

S = $\{(x, y, z) \in R^3 | (x, y, z) \neq (0, 0, 0)\}$

2

6. $f(x, y, z) = \ln(4 - x^2 - y^2 - z^2)$

For the logarithm to be defined and continuous, we require $4 - x^2 - y^2 - z^2 > 0$, which means $x^2 + y^2 + z^2 < 4$. This represents the open ball centered at the origin with radius

2. S = $\{(x, y, z) \in R^3 | x^2 + y^2 + z^2 < 4\}$

3. Final Answer

1

7. S = $R^2$

1

8. S = $R^2$

1

9. S = $\{(x, y) \in R^2 | x^2 + y^2 < 1\}$

2

0. S = $\{(x, y) \in R^2 | y > -x - 1\}$

2

1. S = $\{(x, y) \in R^2 | y \neq x^2\}$

2

2. S = $R^2$

2

3. S = $\{(x, y) \in R^2 | y \leq x + 1\}$

2

4. S = $\{(x, y) \in R^2 | x^2 + y^2 < 4\}$

2

5. S = $\{(x, y, z) \in R^3 | (x, y, z) \neq (0, 0, 0)\}$

2

6. S = $\{(x, y, z) \in R^3 | x^2 + y^2 + z^2 < 4\}$