`lim_(xrarr0)(|sinx|)/(x)` is equal to

Options

A. 1
B. `-1`
C. Does not exist
D. None of these

Explanation

To solve the limit problem \( \lim_{x \to 0} \frac{|\sin x|}{x} \), we will evaluate both the left-hand limit and the right-hand limit.

### Step-by-Step Solution:

1. **Identify the limit**: We need to evaluate \( \lim_{x \to 0} \frac{|\sin x|}{x} \).

2. **Consider the left-hand limit**:
\[ \lim_{x \to 0^-} \frac{|\sin x|}{x} \]
When \( x \) approaches 0 from the left (i.e., \( x < 0 \)), \( \sin x \) is negative. Therefore, \( |\sin x| = -\sin x \). Thus, we can rewrite the limit as:
\[ \lim_{x \to 0^-} \frac{-\sin x}{x} \]
We can factor out the negative sign:
\[ -\lim_{x \to 0^-} \frac{\sin x}{x} \]
We know from standard limit results that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Therefore:
\[ -\lim_{x \to 0^-} \frac{\sin x}{x} = -1 \]

3. **Consider the right-hand limit**:
\[ \lim_{x \to 0^+} \frac{|\sin x|}{x} \]
When \( x \) approaches 0 from the right (i.e., \( x > 0 \)), \( \sin x \) is positive. Therefore, \( |\sin x| = \sin x \). Thus, we can rewrite the limit as:
\[ \lim_{x \to 0^+} \frac{\sin x}{x} \]
Again, using the standard limit result:
\[ \lim_{x \to 0^+} \frac{\sin x}{x} = 1 \]

4. **Compare the limits**:
- Left-hand limit: \( -1 \)
- Right-hand limit: \( 1 \)

Since the left-hand limit and the right-hand limit are not equal, we conclude that the limit does not exist.

### Final Answer:
\[ \lim_{x \to 0} \frac{|\sin x|}{x} \text{ does not exist.} \]