`Lt_(x to pi)(sinx)/(x-pi)` is equal to

Options

A. A. 0
B. B. 1
C. C. `-1`
D. D. none of these

Answer

To solve the limit \( \lim_{x \to \pi} \frac{\sin x}{x - \pi} \), we can use L'Hôpital's Rule since substituting \( x = \pi \) gives us the indeterminate form \( \frac{0}{0} \).

### Step-by-step Solution:

1. **Identify the limit**:
\[ \lim_{x \to \pi} \frac{\sin x}{x - \pi} \]
When we substitute \( x = \pi \):
\[ \sin(\pi) = 0 \quad \text{and} \quad \pi - \pi = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

**Hint**: Check the values of the numerator and denominator at the limit point to see if it's an indeterminate form.

2. **Apply L'Hôpital's Rule**:
Since we have \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
Here, \( f(x) = \sin x \) and \( g(x) = x - \pi \).

Differentiate the numerator and the denominator:
\[ f'(x) = \cos x \quad \text{and} \quad g'(x) = 1 \]

**Hint**: Remember to differentiate both the numerator and the denominator separately.

3. **Evaluate the new limit**:
Now we can rewrite the limit:
\[ \lim_{x \to \pi} \frac{\cos x}{1} \]
Substitute \( x = \pi \):
\[ \cos(\pi) = -1 \]
Therefore:
\[ \lim_{x \to \pi} \frac{\sin x}{x - \pi} = -1 \]

**Hint**: After applying L'Hôpital's Rule, always substitute back into the limit to find the final value.

4. **Conclusion**:
The limit \( \lim_{x \to \pi} \frac{\sin x}{x - \pi} \) is equal to \( -1 \).

**Final Answer**: The correct option is \( c) -1 \).