To evaluate the limit

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

using L'Hôpital's Rule, we first check if the limit has an indeterminate form.

Step 1: Check for Indeterminate Form

Substituting $x = -2$ in the expression:

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

This is an indeterminate form, so we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we take the derivative of the numerator and the denominator separately.

1. Numerator: The derivative of $x + 2$ is $1$.
2. Denominator: The derivative of $x^2 - 4$ is $2x$.

So, we can rewrite the limit as:

$\lim_{x \to -2} \frac{1}{2x}$

Step 3: Substitute $x = -2$ Again

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

$\frac{1}{2(-2)} = \frac{1}{-4} = -\frac{1}{4}$

Final Answer

$\lim_{x \to -2} \frac{x + 2}{x^2 - 4} = -\frac{1}{4}$

Would you like further details on any part of this solution?

Here are some related questions for further exploration:

1. How is L'Hôpital's Rule derived?
2. What are other examples where L'Hôpital's Rule is used for limits?
3. How can you determine when not to use L'Hôpital's Rule?
4. How do you handle higher-order derivatives if the $0/0$ form persists?
5. Can L'Hôpital's Rule be applied to infinite limits?

Tip: Always confirm that a limit is in an indeterminate form (like $0/0$ or $\infty/\infty$) before applying L'Hôpital's Rule.