We are asked to evaluate the following limit: $\lim_{x \to 1} \frac{x^3 - 2x + 1}{x^2 + 4x - 5}$.

1. Problem Description

We are asked to evaluate the following limit:

$\lim_{x \to 1} \frac{x^3 - 2x + 1}{x^2 + 4x - 5}$.

2. Solution Steps

First, we substitute $x = 1$ into the expression:

$\frac{1^3 - 2(1) + 1}{1^2 + 4(1) - 5} = \frac{1 - 2 + 1}{1 + 4 - 5} = \frac{0}{0}$.

Since we have an indeterminate form $\frac{0}{0}$, we need to factor the numerator and the denominator.

Let's factor the numerator $x^3 - 2x + 1$. Since the numerator is 0 when $x=1$, we know that $(x-1)$ is a factor. We can perform polynomial long division or synthetic division to find the other factor.

Dividing $x^3 - 2x + 1$ by $(x-1)$ gives:

$x^3 - 2x + 1 = (x-1)(x^2 + x - 1)$.

Now let's factor the denominator $x^2 + 4x - 5$. We look for two numbers that multiply to $-5$ and add to $4$. These numbers are $5$ and $-1$. So,

$x^2 + 4x - 5 = (x+5)(x-1)$.

Now we can rewrite the limit as:

$\lim_{x \to 1} \frac{(x-1)(x^2 + x - 1)}{(x+5)(x-1)}$.

We can cancel the $(x-1)$ terms:

$\lim_{x \to 1} \frac{x^2 + x - 1}{x+5}$.

Now we can substitute $x = 1$ into the simplified expression:

$\frac{1^2 + 1 - 1}{1 + 5} = \frac{1 + 1 - 1}{6} = \frac{1}{6}$.

3. Final Answer

$\frac{1}{6}$