`lim_(xto0)(|sin x|)/x` is equal to :

Options

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

Answer

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

### Step 1: Define the limit
Let \( L = \lim_{x \to 0} \frac{|\sin x|}{x} \).

### Step 2: Analyze the right-hand limit
We first consider the right-hand limit as \( x \) approaches 0 from the positive side (i.e., \( x \to 0^+ \)):
\[ \lim_{x \to 0^+} \frac{|\sin x|}{x} = \lim_{x \to 0^+} \frac{\sin x}{x} \]
Since \( \sin x \) is positive for \( x > 0 \), we can write:
\[ \lim_{x \to 0^+} \frac{\sin x}{x} = 1 \]

### Step 3: Analyze the left-hand limit
Next, we consider the left-hand limit as \( x \) approaches 0 from the negative side (i.e., \( x \to 0^- \)):
\[ \lim_{x \to 0^-} \frac{|\sin x|}{x} = \lim_{x \to 0^-} \frac{-\sin x}{x} \]
In this case, since \( \sin x \) is negative for \( x < 0 \), we have:
\[ \lim_{x \to 0^-} \frac{-\sin x}{x} = -\lim_{x \to 0^-} \frac{\sin x}{x} \]
Again, we find that:
\[ \lim_{x \to 0^-} \frac{\sin x}{x} = 1 \quad \text{(as \( x \) approaches 0)} \]
Thus,
\[ \lim_{x \to 0^-} \frac{-\sin x}{x} = -1 \]

### Step 4: Compare the limits
Now we compare the right-hand limit and the left-hand limit:
- Right-hand limit: \( \lim_{x \to 0^+} \frac{|\sin x|}{x} = 1 \)
- Left-hand limit: \( \lim_{x \to 0^-} \frac{|\sin x|}{x} = -1 \)

Since the right-hand limit (1) is not equal to the left-hand limit (-1), we conclude that the limit does not exist.

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