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

Options

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

Explanation

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 the value of \( x \)

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

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

### Step 2: Calculate \( \sin^2\left(\frac{\pi}{6}\right) \)

Now, we calculate \( \sin^2\left(\frac{\pi}{6}\right) \):

\[ \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}. \]

### Step 3: Substitute into the numerator

Now substitute \( \sin\left(\frac{\pi}{6}\right) \) and \( \sin^2\left(\frac{\pi}{6}\right) \) into the numerator:

\[ 2\sin^2\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right) - 1 = 2\left(\frac{1}{4}\right) + \frac{1}{2} - 1. \]

Calculating this gives:

\[ = \frac{1}{2} + \frac{1}{2} - 1 = 0. \]

### Step 4: Substitute into the denominator

Now substitute into the denominator:

\[ 2\sin^2\left(\frac{\pi}{6}\right) - 3\sin\left(\frac{\pi}{6}\right) + 1 = 2\left(\frac{1}{4}\right) - 3\left(\frac{1}{2}\right) + 1. \]

Calculating this gives:

\[ = \frac{1}{2} - \frac{3}{2} + 1 = 0. \]

### Step 5: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule:

\[ \lim_{x \to \frac{\pi}{6}} \frac{2\sin^2 x + \sin x - 1}{2\sin^2 x - 3\sin x + 1} = \lim_{x \to \frac{\pi}{6}} \frac{d}{dx}(2\sin^2 x + \sin x - 1) \bigg/ \frac{d}{dx}(2\sin^2 x - 3\sin x + 1). \]

### Step 6: Differentiate the numerator and denominator

1. Differentiate the numerator:

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

2. Differentiate the denominator:

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

### Step 7: Substitute \( x = \frac{\pi}{6} \) again

Now we substitute \( x = \frac{\pi}{6} \) again:

Numerator:

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

Denominator:

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

### Step 8: Final limit calculation

Now we can compute the limit:

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

Thus, the limit is

\[ \boxed{-3}. \]