Certainly! Let's solve the problem using **L'Hôpital's Rule** step by step.

### Problem:
We want to evaluate the following limit using L'Hôpital's Rule:

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

### Step 1: Combine the Fractions
First, we'll combine the two fractions into a single fraction:

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

Simplify the numerator:

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

Thus, the limit becomes:

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

### Step 2: Check the Form
Substituting \(x = 2\) gives:

\[
\frac

Learn how to evaluate the limit lim(x → 2) [4/(x^2 - 4) - 1/(x - 2)] using L'Hôpital's Rule. Step-by-step solution with detailed fraction manipulation and substitution. Suitable for advanced high school students and college-level mathematics enthusiasts.

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