Examine if `sin|x|` is a continuous function.

Answer

To determine if the function \( f(x) = \sin|x| \) is continuous, we will follow these steps:

### Step 1: Understand the function
The function \( f(x) = \sin|x| \) involves the sine of the absolute value of \( x \). The absolute value function \( |x| \) is defined as:
\[ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \]
This means that \( f(x) \) will behave differently for positive and negative values of \( x \).

### Step 2: Analyze the function for positive and negative values
For \( x \geq 0 \):
\[ f(x) = \sin(x) \]
For \( x < 0 \):
\[ f(x) = \sin(-x) = -\sin(x) \]
Thus, we can rewrite the function as:
\[ f(x) = \begin{cases} \sin(x) & \text{if } x \geq 0 \\ -\sin(x) & \text{if } x < 0 \end{cases} \]

### Step 3: Check continuity at \( x = 0 \)
To check if \( f(x) \) is continuous at \( x = 0 \), we need to verify that:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

1. **Calculate \( f(0) \)**:
\[ f(0) = \sin|0| = \sin(0) = 0 \]

2. **Calculate the left-hand limit**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -\sin(x) = -\sin(0) = 0 \]

3. **Calculate the right-hand limit**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sin(x) = \sin(0) = 0 \]

### Step 4: Conclusion
Since:
\[ \lim_{x \to 0^-} f(x) = 0, \quad \lim_{x \to 0^+} f(x) = 0, \quad \text{and} \quad f(0) = 0 \]
we have:
\[ \lim_{x \to 0} f(x) = f(0) \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 5: Check continuity for all \( x \)
The functions \( \sin(x) \) and \( -\sin(x) \) are both continuous for all \( x \). Since \( f(x) \) is defined piecewise with continuous functions on both sides of \( x = 0 \) and is continuous at \( x = 0 \), we conclude that \( f(x) = \sin|x| \) is continuous for all \( x \).

### Final Answer
The function \( f(x) = \sin|x| \) is continuous for all \( x \).

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