`lim_(x to pi//6) (2sin^2x+sinx-1)/(2sin^2x-3sinx+1)=`

Options

A. 3
B. -3
C. 6
D. 0

Explanation

To solve 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, let's substitute \( x = \frac{\pi}{6} \) into the expression to check if we get a determinate form.

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

Now, substituting into 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 are 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Factor the numerator and denominator

Next, we need to factor both the numerator and the denominator.

**Numerator:**
\[ 2\sin^2 x + \sin x - 1 = (2\sin x - 1)(\sin x + 1) \]

**Denominator:**
\[ 2\sin^2 x - 3\sin x + 1 = (2\sin x - 1)(\sin x - 1) \]

### Step 3: Rewrite the limit

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{6}} \frac{(2\sin x - 1)(\sin x + 1)}{(2\sin x - 1)(\sin x - 1)} \]

### Step 4: Cancel common factors

We can cancel \( (2\sin x - 1) \) from the numerator and denominator (as long as \( \sin x \neq \frac{1}{2} \)):

\[ \lim_{x \to \frac{\pi}{6}} \frac{\sin x + 1}{\sin x - 1} \]

### Step 5: Substitute \( x = \frac{\pi}{6} \) again

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

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

So we have:

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

### Final Answer

Thus, the limit is:

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