We are tasked with evaluating the following limit:

$\lim_{x \to -2} \frac{x^2 + 4x + 4}{x^4 - 16}$

Step 1: Simplify the expression

We start by factoring both the numerator and denominator.

Numerator:

$x^2 + 4x + 4 = (x + 2)^2$ (The numerator is a perfect square.)

Denominator:

$x^4 - 16 = (x^2 - 4)(x^2 + 4)$ Using the difference of squares, we can further factor $x^2 - 4$: $x^2 - 4 = (x - 2)(x + 2)$ Thus, the denominator becomes: $x^4 - 16 = (x - 2)(x + 2)(x^2 + 4)$

Step 2: Rewrite the limit expression

Now, we can rewrite the entire expression: $\frac{(x + 2)^2}{(x - 2)(x + 2)(x^2 + 4)}$

Step 3: Cancel common factors

Notice that both the numerator and denominator contain a factor of $(x + 2)$. We can cancel out one factor of $(x + 2)$ (as long as $x \neq -2$):

$\frac{(x + 2)}{(x - 2)(x^2 + 4)}$

Step 4: Evaluate the limit

Now, we can substitute $x = -2$ into the simplified expression:

$\frac{(-2 + 2)}{(-2 - 2)((-2)^2 + 4)} = \frac{0}{(-4)(4 + 4)} = \frac{0}{-32} = 0$

Final Answer:

$\lim_{x \to -2} \frac{x^2 + 4x + 4}{x^4 - 16} = 0$

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Would you like further clarification or details on any of the steps? Here are some related questions you might find helpful:

1. What happens if you evaluate the limit as $x \to 2$ for this expression?
2. How do you handle limits involving indeterminate forms like $\frac{0}{0}$?
3. What other factoring techniques are commonly used in limits?
4. How can you apply L'Hopital's Rule to this problem?
5. What is the significance of canceling common factors in rational functions?

Tip: Always look for common factors when simplifying rational expressions in limits, as this can often help resolve indeterminate forms like $\frac{0}{0}$.