The limit you're trying to evaluate is:

$\lim_{x \to 1} \frac{x^3 - 4x^2 + 2x + 1}{x^2 - 1}$

To solve this, let's follow these steps:

Step 1: Factor the denominator

The denominator $x^2 - 1$ is a difference of squares, so we can factor it as: $x^2 - 1 = (x - 1)(x + 1)$

Step 2: Plug $x = 1$ into the expression

Direct substitution of $x = 1$ into the expression gives: $\frac{1^3 - 4(1)^2 + 2(1) + 1}{1^2 - 1} = \frac{1 - 4 + 2 + 1}{1 - 1} = \frac{0}{0}$ This results in the indeterminate form $\frac{0}{0}$, so we need to simplify the expression.

Step 3: Factor the numerator

The numerator is $x^3 - 4x^2 + 2x + 1$. We need to attempt factoring it. Using synthetic division or polynomial division, we can divide the numerator by $(x - 1)$, since $x = 1$ is a root (as shown by the substitution).

After factoring, the numerator simplifies as: $x^3 - 4x^2 + 2x + 1 = (x - 1)(x^2 - 3x - 1)$

Step 4: Simplify the expression

Now, the original expression becomes: $\frac{(x - 1)(x^2 - 3x - 1)}{(x - 1)(x + 1)}$

We can cancel out the common factor of $(x - 1)$ from the numerator and denominator (as long as $x \neq 1$): $\frac{x^2 - 3x - 1}{x + 1}$

Step 5: Substitute $x = 1$

Now, substitute $x = 1$ into the simplified expression: $\frac{1^2 - 3(1) - 1}{1 + 1} = \frac{1 - 3 - 1}{2} = \frac{-3}{2}$

Final Answer:

$\lim_{x \to 1} \frac{x^3 - 4x^2 + 2x + 1}{x^2 - 1} = \frac{-3}{2}$

Would you like further details on any of the steps?

Here are some related questions for further practice:

1. What is the limit of $\frac{x^2 - 4}{x - 2}$ as $x \to 2$?
2. How do you factor a cubic polynomial?
3. How do you identify and solve indeterminate forms?
4. What is the difference of squares and how is it applied in limits?
5. What is L'Hopital's Rule, and when can it be applied?

Tip: When you encounter indeterminate forms like $\frac{0}{0}$, try factoring the numerator and denominator or apply L'Hopital’s Rule for simplification.