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The provided image shows the following mathematical expression:

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

Let's solve this limit step by step.

### Step 1: Direct Substitution
First, we'll substitute \( x = 1 \) directly into the expression to see if we get a determinate value.

\[
\text{Numerator at } x = 1: 2(1)^4 - 2(1)^3 - (1)^2 + 1 = 2 - 2 - 1 + 1 = 0
\]
\[
\text{Denominator at } x = 1: (1)^4 - (1)^2 - 2(1) + 2 = 1 - 1 - 2 + 2 = 0
\]

Since both the numerator and the denominator are \( 0 \) when \( x = 1 \), we have a \( \frac{0}{0} \) indeterminate form. Therefore, we can apply **L'Hôpital's Rule**.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that for limits of the form \( \frac{0}{0} \), we can differentiate the numerator and the denominator separately and then take the limit.

#### Differentiating the numerator:
\[
\frac{d}{dx} \left( 2x^4 - 2x^3 - x^2 + 1 \right) = 8x^3 - 6x^2 - 2x
\]

#### Differentiating the denominator:
\[
\frac{d}{dx} \left( x^4 - x^2 - 2x + 2 \right) = 4x^3 - 2x - 2
\]

### Step 3: Take the limit again
Now, we take the limit as \( x \to 1 \) of the new expression:

\[
\lim_{{x \to 1}} \frac{8x^3 - 6x^2 - 2x}{4x^3 - 2x - 2}
\]

Substitute \( x = 1 \):

\[
\text{Numerator: } 8(1)^3 - 6(1)^2 - 2(1) = 8 - 6 - 2 = 0
\]
\[
\text{Denominator: } 4(1)^3 - 2(1) - 2 = 4 - 2 - 2 = 0
\]

We still have a \( \frac{0}{0} \) form, so we apply **L'Hôpital's Rule** again.

### Step 4: Apply L'Hôpital's Rule again
Differentiate the numerator and denominator again.

#### Second derivative of the numerator:
\[
\frac{d}{dx} (8x^3 - 6x^2 - 2x) = 24x^2 - 12x - 2
\]

#### Second derivative of the denominator:
\[
\frac{d}{dx} (4x^3 - 2x - 2) = 12x^2 - 2
\]

### Step 5: Take the limit one more time
Now, take the limit as \( x \to 1 \):

\[
\lim_{{x \to 1}} \frac{24x^2 - 12x - 2}{12x^2 - 2}
\]

Substitute \( x = 1 \):

\[
\text{Numerator: } 24(1)^2 - 12(1) - 2 = 24 - 12 - 2 = 10
\]
\[
\text{Denominator: } 12(1)^2 - 2 = 12 - 2 = 10
\]

Thus, the limit is:

\[
\boxed{1}
\]

### Final Answer:
The value of the limit is \( 1 \).

Would you like further details or have any questions?

Here are 5 related questions to explore:
1. How does L'Hôpital's Rule work and when can it be applied?
2. Can L'Hôpital's Rule be applied multiple times?
3. How do you differentiate polynomials step-by-step?
4. What other techniques are available for solving indeterminate forms?
5. How can we evaluate limits without using L'Hôpital's Rule?

**Tip**: Always check if direct substitution works first when solving limits before applying advanced methods like L'Hôpital's Rule.

Lim (2x^4 - 2x^3 - x^2 + 1) / (x^4 - x^2 - 2x + 2) as x approaches 1

Solve the limit problem Lim (2x^4 - 2x^3 - x^2 + 1) / (x^4 - x^2 - 2x + 2) as x approaches 1 using L'Hôpital's Rule. Explore indeterminate forms, polynomial differentiation, and advanced calculus techniques.

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