`lim_(x to pi)(sin 3x)/(sin 2x)` is equal to

Options

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

Answer

To solve the limit \( \lim_{x \to \pi} \frac{\sin 3x}{\sin 2x} \), we will follow these steps:

### Step 1: Substitute \( x = \pi \)
First, we substitute \( x = \pi \) into the expression:

\[ \sin(3\pi) = 0 \quad \text{and} \quad \sin(2\pi) = 0 \]

Thus, we have:

\[ \frac{\sin(3\pi)}{\sin(2\pi)} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator:

\[ \text{Numerator: } \frac{d}{dx}(\sin 3x) = 3\cos 3x \]
\[ \text{Denominator: } \frac{d}{dx}(\sin 2x) = 2\cos 2x \]

Now we can rewrite the limit:

\[ \lim_{x \to \pi} \frac{\sin 3x}{\sin 2x} = \lim_{x \to \pi} \frac{3\cos 3x}{2\cos 2x} \]

### Step 3: Substitute \( x = \pi \) Again
Now we substitute \( x = \pi \) into the new expression:

\[ \cos(3\pi) = -1 \quad \text{and} \quad \cos(2\pi) = 1 \]

So we have:

\[ \lim_{x \to \pi} \frac{3\cos 3x}{2\cos 2x} = \frac{3(-1)}{2(1)} = \frac{-3}{2} \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to \pi} \frac{\sin 3x}{\sin 2x} = -\frac{3}{2} \]

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