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The given limit is:

\[
\lim_{{x \to 2}} \frac{x^4 - 8x^2 + 16}{x^2 - 4x + 4}.
\]

Let's evaluate this step by step:

### Step 1: Simplify the denominator

The denominator is a quadratic expression:
\[
x^2 - 4x + 4 = (x - 2)^2.
\]

### Step 2: Analyze the numerator

The numerator is a quartic expression:
\[
x^4 - 8x^2 + 16.
\]

We can try factoring this expression. Notice that it can be rewritten as a quadratic in \(x^2\):
\[
(x^2)^2 - 8(x^2) + 16.
\]

This is a perfect square trinomial:
\[
x^4 - 8x^2 + 16 = (x^2 - 4)^2.
\]

### Step 3: Rewrite the limit

Now the expression becomes:
\[
\lim_{{x \to 2}} \frac{(x^2 - 4)^2}{(x - 2)^2}.
\]

### Step 4: Simplify

Since \(x^2 - 4 = (x - 2)(x + 2)\), we can substitute that into the expression:
\[
\lim_{{x \to 2}} \frac{((x - 2)(x + 2))^2}{(x - 2)^2}.
\]

Simplifying:
\[
\lim_{{x \to 2}} \frac{(x - 2)^2(x + 2)^2}{(x - 2)^2}.
\]

Cancel the \((x - 2)^2\) terms:
\[
\lim_{{x \to 2}} (x + 2)^2.
\]

### Step 5: Evaluate the limit

Substitute \(x = 2\) into \((x + 2)^2\):
\[
(2 + 2)^2 = 4^2 = 16.
\]

Thus, the value of the limit is:
\[
\boxed{16}.
\]

Would you like further clarification or more details on this solution?

Here are some related questions for practice:
1. What is the factorization process for quartic expressions like \(x^4 - 8x^2 + 16\)?
2. How do you approach limits involving indeterminate forms like \(\frac{0}{0}\)?
3. Can this problem be solved using L'Hopital's Rule?
4. What is the general process for factoring perfect square trinomials?
5. How would the limit change if the denominator were \(x^2 + 4x + 4\) instead?

**Tip:** When encountering limits involving factored polynomials, always check for common factors in the numerator and denominator that may cancel out.

Find the limit of (x^4 - 8x^2 + 16) / (x^2 - 4x + 4) as x approaches 2.

Learn how to solve the limit of (x^4 - 8x^2 + 16) / (x^2 - 4x + 4) as x approaches 2. This step-by-step solution includes factoring perfect square trinomials and simplifying indeterminate forms. Suitable for high school students in Grades 10-12.

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