Let's solve the given limit step by step:

The limit is:

$\lim_{x \to 1} \frac{2x^2 - x - 1}{3x^2 - x - 2}$

Step 1: Factorize the numerator and denominator

1. Numerator: $2x^2 - x - 1$
   To factorize $2x^2 - x - 1$, we split the middle term: $2x^2 - x - 1 = (2x + 1)(x - 1)$

2. Denominator: $3x^2 - x - 2$
   Similarly, factorizing $3x^2 - x - 2$: $3x^2 - x - 2 = (3x + 2)(x - 1)$

Step 2: Simplify the expression

The given limit now becomes: $\lim_{x \to 1} \frac{(2x + 1)(x - 1)}{(3x + 2)(x - 1)}$

Since $x - 1$ appears in both numerator and denominator, and $x \to 1$ (but $x \neq 1$), we can cancel the $x - 1$ terms:

$\lim_{x \to 1} \frac{2x + 1}{3x + 2}$

Step 3: Substitute $x = 1$

Now, substitute $x = 1$ into the simplified expression:

$\frac{2(1) + 1}{3(1) + 2} = \frac{2 + 1}{3 + 2} = \frac{3}{5}$

Final Answer:

The value of the limit is:

$\boxed{\frac{3}{5}}$

This corresponds to option D.

Do you need a detailed explanation for any step? Here are 5 related questions to practice:

1. Find $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.
2. Factorize the polynomial $x^2 + 5x + 6$ and evaluate the limit as $x \to -2$.
3. Simplify $\frac{x^2 - 9}{x - 3}$ and evaluate the limit as $x \to 3$.
4. Solve $\lim_{x \to 2} \frac{3x^2 - 2x - 8}{x - 2}$.
5. Factorize $x^3 - x^2 - x + 1$.

Tip: Always factorize both numerator and denominator when limits result in indeterminate forms like $\frac{0}{0}$.