Evaluate the following limits:
`lim_(xrarr pi//6)((2sin^(2)x+sinx-1))/((2sin^(2)x-3sinx+1))`

Explanation

To evaluate 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 to check if it results in an indeterminate form.

Calculating \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \):

- 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 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: Simplify the limit

Now we can rewrite the limit using the factored forms:

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

We can cancel out the common factor \( (2\sin x - 1) \) (as long as \( \sin x \neq \frac{1}{2} \), which is valid in the limit):

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

### Step 4: Substitute \( x = \frac{\pi}{6} \) again

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

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

### Final Answer

Thus, the limit is

\[ \boxed{-3}. \]