We need to find the limit of the expression $\frac{2x^3 - 2x + 1}{x^2 + 4x - 5}$ as $x$ approaches $1$.

1. Problem Description

We need to find the limit of the expression $\frac{2x^3 - 2x + 1}{x^2 + 4x - 5}$ as $x$ approaches $1$.

2. Solution Steps

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

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

Since we get a $0$ in the denominator, we cannot directly substitute $x = 1$. This implies that we should try to factor the numerator and denominator to see if we can simplify the expression and eliminate the singularity.

However, the numerator $2x^3 - 2x + 1$ does not seem to have an obvious factor of $(x - 1)$. Let's check the denominator:

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

Since the denominator has a factor of $(x - 1)$, let's divide $2x^3 - 2x + 1$ by $(x - 1)$ to see if we can factor the numerator.

Polynomial long division of $(2x^3 - 2x + 1)$ by $(x - 1)$:

```

2x^2 + 2x

x - 1 | 2x^3 + 0x^2 - 2x + 1

2x^3 - 2x^2

-------------

2x^2 - 2x

2x^2 - 2x

-------------

0 + 1

```

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

Therefore, the expression becomes:

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

Since the numerator does not have a factor of $(x - 1)$, the limit does not exist. More specifically, since the denominator approaches $0$ as $x$ approaches $1$ and the numerator approaches $1$, the expression approaches infinity.

Let's analyze the limit as $x$ approaches $1$ from the left and from the right.

As $x \to 1^+$, $x - 1 \to 0^+$ and $x + 5 \to 6$. Thus, the denominator $(x - 1)(x + 5) \to 0^+$.

As $x \to 1^+$, the numerator $2x^3 - 2x + 1 \to 2(1)^3 - 2(1) + 1 = 1$.

Therefore, $\lim_{x \to 1^+} \frac{2x^3 - 2x + 1}{x^2 + 4x - 5} = \lim_{x \to 1^+} \frac{1}{0^+} = +\infty$.

As $x \to 1^-$, $x - 1 \to 0^-$ and $x + 5 \to 6$. Thus, the denominator $(x - 1)(x + 5) \to 0^-$.

As $x \to 1^-$, the numerator $2x^3 - 2x + 1 \to 2(1)^3 - 2(1) + 1 = 1$.

Therefore, $\lim_{x \to 1^-} \frac{2x^3 - 2x + 1}{x^2 + 4x - 5} = \lim_{x \to 1^-} \frac{1}{0^-} = -\infty$.

Since the left-hand limit and the right-hand limit are not equal, the limit does not exist.

3. Final Answer

The limit does not exist.

More specifically, the limit goes to infinity.