`lim_(xrarr0) (sinx)/(x)=`?

Options

A. 0
B. 2
C. `(1)/(2)`
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Identify the Limit**: We need to evaluate the limit as \( x \) approaches 0 for the function \( \frac{\sin x}{x} \).
\[ \lim_{x \to 0} \frac{\sin x}{x} \]

2. **Substitute \( x = 0 \)**: When we substitute \( x = 0 \) into the function, we get:
\[ \frac{\sin(0)}{0} = \frac{0}{0} \]
This is an indeterminate form, so we cannot directly evaluate the limit.

3. **Apply L'Hôpital's Rule**: Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This rule states that if the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator separately.

4. **Differentiate the Numerator and Denominator**:
- The derivative of the numerator \( \sin x \) is \( \cos x \).
- The derivative of the denominator \( x \) is \( 1 \).

Therefore, we rewrite the limit:
\[ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} \]

5. **Evaluate the New Limit**: Now we can substitute \( x = 0 \) into the new limit:
\[ \cos(0) = 1 \]

6. **Final Result**: Thus, the value of the limit is:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

### Conclusion:
The final answer is:
\[ \boxed{1} \]