1. Key Concepts and Formulas

• Definition of Continuity: A function $f(x)$ is continuous at a point $x=c$ if and only if:
  • $f(c)$ is defined.
  • $\lim_{x \to c} f(x)$ exists.
  • $\lim_{x \to c} f(x) = f(c)$.
• Properties of Rational and Irrational Numbers: In any neighborhood of a real number, there are both rational and irrational numbers. Specifically, for any real number $c$, and any $\epsilon > 0$, the interval $(c-\epsilon, c+\epsilon)$ contains both rational and irrational numbers.
• Limit of a function in the neighborhood of a point: For a limit $\lim_{x \to c} f(x)$ to exist, the function must approach the same value as $x$ approaches $c$ from both the left and the right. If the function behaves differently for different types of numbers (rational vs. irrational) in the neighborhood, we need to consider the limits for each type.

1. Step-by-Step Solution

• Step 1: Analyze the function definition. The function $f(x)$ is defined on the interval $[-5, 5]$ as: $f(x) = \begin{cases} x & \text{if } x \in \mathbb{Q} \\ -x & \text{if } x \notin \mathbb{Q} \end{cases}$ This piecewise definition means that the function's value depends on whether the input $x$ is a rational or an irrational number.

• Step 2: Consider continuity at a non-zero rational number $a$. Let $a$ be a rational number such that $a \in [-5, 5]$ and $a \neq 0$. According to the function definition, $f(a) = a$ since $a$ is rational. Now, we need to evaluate the limit $\lim_{x \to a} f(x)$. Since $a$ is a rational number, its immediate neighborhood contains both rational and irrational numbers. If we consider a sequence of rational numbers $x_n \to a$, then $f(x_n) = x_n$, so $\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = a$. If we consider a sequence of irrational numbers $y_n \to a$, then $f(y_n) = -y_n$, so $\lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} (-y_n) = -a$. For the limit $\lim_{x \to a} f(x)$ to exist, the limits along all sequences must be the same. Since $a \neq 0$, we have $a \neq -a$. Therefore, $\lim_{x \to a} f(x)$ does not exist when $a$ is a non-zero rational number. Since the limit does not exist, $f(x)$ is discontinuous at any non-zero rational number $a$.

• Step 3: Consider continuity at a non-zero irrational number $a$. Let $a$ be an irrational number such that $a \in [-5, 5]$ and $a \neq 0$. According to the function definition, $f(a) = -a$ since $a$ is irrational. Now, we need to evaluate the limit $\lim_{x \to a} f(x)$. Since $a$ is an irrational number, its immediate neighborhood contains both rational and irrational numbers. If we consider a sequence of rational numbers $x_n \to a$, then $f(x_n) = x_n$, so $\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = a$. If we consider a sequence of irrational numbers $y_n \to a$, then $f(y_n) = -y_n$, so $\lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} (-y_n) = -a$. For the limit $\lim_{x \to a} f(x)$ to exist, the limits along all sequences must be the same. Since $a \neq 0$, we have $a \neq -a$. Therefore, $\lim_{x \to a} f(x)$ does not exist when $a$ is a non-zero irrational number. Since the limit does not exist, $f(x)$ is discontinuous at any non-zero irrational number $a$.

• Step 4: Consider continuity at $x = 0$. We need to check the three conditions for continuity at $x=0$.

  1. Is $f(0)$ defined? Since $0$ is a rational number ($0 = 0/1$), $f(0) = 0$ according to the definition. So, $f(0)$ is defined.
  2. Does $\lim_{x \to 0} f(x)$ exist? We need to check the limit from the neighborhood of $0$. Consider a sequence of rational numbers $x_n \to 0$. Then $f(x_n) = x_n$, so $\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = 0$. Consider a sequence of irrational numbers $y_n \to 0$. Then $f(y_n) = -y_n$, so $\lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} (-y_n) = 0$. Since the limit of $f(x)$ as $x$ approaches $0$ is $0$ regardless of whether $x$ is rational or irrational, the limit $\lim_{x \to 0} f(x)$ exists and is equal to $0$.
  3. Is $\lim_{x \to 0} f(x) = f(0)$? We found that $\lim_{x \to 0} f(x) = 0$ and $f(0) = 0$. Therefore, $\lim_{x \to 0} f(x) = f(0)$.

  Since all three conditions are met, $f(x)$ is continuous at $x=0$.

• Step 5: Conclude about the continuity of $f(x)$ over the interval $[-5, 5]$. From Steps 2 and 3, we established that $f(x)$ is discontinuous at every non-zero rational number and every non-zero irrational number within the interval $[-5, 5]$. From Step 4, we established that $f(x)$ is continuous at $x=0$. Therefore, $f(x)$ is continuous at $x=0$ and discontinuous at every other point in the interval $[-5, 5]$. This means $f(x)$ is continuous at every $x$ except $x=0$.

1. Common Mistakes & Tips

• Confusing limit existence with continuity: Remember that for continuity, the limit must exist AND be equal to the function's value at that point.
• Ignoring the behavior of the function for both rational and irrational inputs: When dealing with piecewise functions defined on rational/irrational numbers, it's crucial to check the limit by approaching the point from sequences of both types of numbers.
• Special case of $x=0$: The point $x=0$ is often a special case in such problems because both $x$ and $-x$ are $0$ at $x=0$, thus resolving the conflict between the two definitions.

1. Summary

The function $f(x)$ is defined as $x$ for rational numbers and $-x$ for irrational numbers. To determine continuity, we examined the behavior of the function at different points. For any non-zero rational or irrational number $a$, the limit of $f(x)$ as $x$ approaches $a$ does not exist because the function approaches different values for rational and irrational numbers in the neighborhood of $a$. However, at $x=0$, $f(0)=0$. The limit as $x$ approaches $0$ from both rational and irrational sequences is $0$, and this equals $f(0)$. Thus, the function is continuous only at $x=0$ and discontinuous everywhere else in its domain $[-5, 5]$.

1. Final Answer

The final answer is $\boxed{A}$ which corresponds to option (A).