What is `lim_(xto(pi)/(6))(2sin^(2)x+sinx-1)/(2sin^(2)x-3sinx+1)` to ?

Options

A. A. `-(1)/(2)`
B. B. `-(1)/(3)`
C. C. `-2`
D. D. `-3`

Answer

To find the limit

\[ \lim_{x \to \frac{\pi}{6}} \frac{2\sin^2 x + \sin x - 1}{2\sin^2 x - 3\sin x + 1}, \]

we will follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{6} \)

First, we substitute \( x = \frac{\pi}{6} \) into the expression.

\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

Now we calculate the numerator and denominator:

**Numerator:**
\[ 2\sin^2\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right) - 1 = 2\left(\frac{1}{2}\right)^2 + \frac{1}{2} - 1 = 2 \cdot \frac{1}{4} + \frac{1}{2} - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0 \]

**Denominator:**
\[ 2\sin^2\left(\frac{\pi}{6}\right) - 3\sin\left(\frac{\pi}{6}\right) + 1 = 2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 1 = 2 \cdot \frac{1}{4} - \frac{3}{2} + 1 = \frac{1}{2} - \frac{3}{2} + 1 = 0 \]

Since both the numerator and denominator evaluate to 0, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

**Numerator Derivative:**
\[ \frac{d}{dx}(2\sin^2 x + \sin x - 1) = 4\sin x \cos x + \cos x = \cos x(4\sin x + 1) \]

**Denominator Derivative:**
\[ \frac{d}{dx}(2\sin^2 x - 3\sin x + 1) = 4\sin x \cos x - 3\cos x = \cos x(4\sin x - 3) \]

### Step 3: Substitute \( x = \frac{\pi}{6} \) again

Now we substitute \( x = \frac{\pi}{6} \) into the derivatives:

**Numerator:**
\[ \cos\left(\frac{\pi}{6}\right)(4\sin\left(\frac{\pi}{6}\right) + 1) = \frac{\sqrt{3}}{2}\left(4 \cdot \frac{1}{2} + 1\right) = \frac{\sqrt{3}}{2}(2 + 1) = \frac{\sqrt{3}}{2} \cdot 3 = \frac{3\sqrt{3}}{2} \]

**Denominator:**
\[ \cos\left(\frac{\pi}{6}\right)(4\sin\left(\frac{\pi}{6}\right) - 3) = \frac{\sqrt{3}}{2}\left(4 \cdot \frac{1}{2} - 3\right) = \frac{\sqrt{3}}{2}(2 - 3) = \frac{\sqrt{3}}{2}(-1) = -\frac{\sqrt{3}}{2} \]

### Step 4: Calculate the limit

Now we can compute the limit:

\[ \lim_{x \to \frac{\pi}{6}} \frac{3\sqrt{3}/2}{-\sqrt{3}/2} = \frac{3\sqrt{3}/2}{-\sqrt{3}/2} = -3 \]

### Final Answer

Thus, the limit is

\[ \boxed{-3}. \]