Certainly! Let's solve the given limit using **L'Hôpital's Rule**.

The limit we need to solve is:

\[
\lim_{x \to 2} \left(\frac{4}{x^2 - 4} - \frac{1}{x - 2}\right)
\]

### Step 1: Combine the Fractions

First, let's write the expression as a single fraction:

\[
\lim_{x \to 2} \frac{4 - (x^2 - 4)}{(x^2 - 4)(x - 2)}
\]

Simplify the numerator:

\[
4 - (x^2 - 4) = 4 - x^2 + 4 = 8 - x^2
\]

So the limit becomes:

\[
\lim_{x \to 2} \frac{8 - x^2}{(x^2 - 4)(x - 2)}
\]

### Step 2: Evaluate the Limit Directly

Now, substitute \(x = 2\) to check if we get an indeterminate form:

- The numerator: \(8 - x^2 = 8 - 4 = 4\)
- The denominator: \(x^2 - 4 = 4 - 4 = 0\) and \(x - 2 = 0\)

The expression \(0/0\) is indeed an indeterminate form, so we can apply **L'Hôpital's Rule**.

### Step 3: Apply L'Hôpital's Rule

L'Hôpital's Rule tells us to differentiate the numerator and the denominator separately:

- Differentiate the numerator \(8 - x^2\) with respect to \(x\):

\[
\frac{d}{dx}(8 - x^2) = -2x
\]

- Differentiate the denominator \((x^2 - 4)(x - 2)\). We'll use the product rule for this:

Let \(u(x) = x^2 - 4\) and \(v(x) = x - 2\).

\[
\frac{d}{dx}[(x^2 - 4)(x - 2)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

Where:

- \(u'(x) = 2x\)
- \(v'(x) = 1\)

Substituting:

\[
\text{Derivative} = (2x)(x - 2) + (x^2 - 4)(1)
\]

Simplifying:

\[
= 2x(x - 2) + x^2 - 4 = 2x^2 - 4x + x^2 - 4 = 3x^2 - 4x - 4
\]

So, applying L'Hôpital's Rule gives us:

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

### Step 4: Substitute \(x = 2\)

Now, substitute \(x = 2\) into the derivative:

- Numerator: \(-2(2) = -4\)
- Denominator: \(3(2)^2 - 4(2) - 4 = 3(4) - 8 - 4 = 12 - 8 - 4 = 0\)

Since we again have an indeterminate form \(\frac{-4}{0}\), L'Hôpital's Rule applies a second time. 

### Step 5: Apply L'Hôpital's Rule Again

Differentiate the numerator and denominator again:

- Numerator: \(\frac{d}{dx}(-2x) = -2\)
- Denominator: \(\frac{d}{dx}(3x^2 - 4x - 4) = 6x - 4\)

So, we get:

\[
\lim_{x \to 2} \frac{-2}{6x - 4}
\]

### Step 6: Evaluate the Limit

Substitute \(x = 2\) into this expression:

\[
\lim_{x \to 2} \frac{-2}{6(2) - 4} = \frac{-2}{12 - 4} = \frac{-2}{8} = \boxed{-\frac{1}{4}}
\]

Thus, the limit is indeed \(\boxed{-\frac{1}{4}}\) using L'Hôpital's Rule.

Would you like any more details or have any questions?

Here are some related questions you might consider:
1. When should you use L'Hôpital's Rule versus simplifying the expression directly?
2. How can you differentiate the product of two functions?
3. What are some other common indeterminate forms where L'Hôpital's Rule applies?
4. How does L'Hôpital's Rule apply to limits involving infinity?
5. Can L'Hôpital's Rule be used for one-sided limits?

**Tip:** L'Hôpital's Rule is especially powerful when dealing with indeterminate forms like \(0/0\) or \(\infty/\infty\). However, always check if the expression can be simplified before applying the rule.

Learn how to solve a limit using L'Hôpital's Rule with a step-by-step solution example. This problem involves differentiation and addresses indeterminate forms in limits. Suitable for advanced high school and college-level students.

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