Evaluate: `lim_(x to 0) (sin x)/(tan x)`

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To evaluate the limit \( \lim_{x \to 0} \frac{\sin x}{\tan x} \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the tangent function**:
We know that \( \tan x = \frac{\sin x}{\cos x} \). Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\sin x}{\tan x} = \lim_{x \to 0} \frac{\sin x}{\frac{\sin x}{\cos x}} = \lim_{x \to 0} \frac{\sin x \cdot \cos x}{\sin x} \]

2. **Simplify the expression**:
As long as \( \sin x \neq 0 \) (which is true for \( x \) approaching 0 but not equal to 0), we can cancel \( \sin x \) from the numerator and denominator:
\[ \lim_{x \to 0} \frac{\sin x \cdot \cos x}{\sin x} = \lim_{x \to 0} \cos x \]

3. **Evaluate the limit**:
Now, we can directly substitute \( x = 0 \) into \( \cos x \):
\[ \cos(0) = 1 \]

4. **Conclusion**:
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{\sin x}{\tan x} = 1 \]

### Final Answer:
\[ \lim_{x \to 0} \frac{\sin x}{\tan x} = 1 \]