Select the correct alternatives out of given four alternatives in each.
`lim_(x to pi)(sinx)/(x-pi)` is-

Options

A. 1
B. 2
C. `-1`
D. does not exists

Answer

To find the limit \( \lim_{x \to \pi} \frac{\sin x}{x - \pi} \), we will follow these steps:

### Step 1: Identify the form of the limit
We start by substituting \( x = \pi \) into the expression:
\[ \frac{\sin(\pi)}{\pi - \pi} = \frac{0}{0} \]
This is an indeterminate form.

**Hint:** When you encounter a \( \frac{0}{0} \) form, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
Here, \( f(x) = \sin x \) and \( g(x) = x - \pi \).

**Hint:** Differentiate the numerator and the denominator separately.

### Step 3: Differentiate the numerator and denominator
Now we differentiate:
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( x - \pi \) is \( 1 \).

So, we rewrite the limit as:
\[ \lim_{x \to \pi} \frac{\cos x}{1} \]

**Hint:** Substitute \( x = \pi \) again after differentiating.

### Step 4: Substitute \( x = \pi \)
Now we substitute \( x = \pi \) into the new limit:
\[ \frac{\cos(\pi)}{1} = \frac{-1}{1} = -1 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to \pi} \frac{\sin x}{x - \pi} = -1 \]

**Final Answer:** The correct alternative is option 3, which is \(-1\).

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