Evaluate `lim_(x to 2) (x^(2) - 4) [(1)/(x + 2) + (1)/(x - 2)]`

Answer

To evaluate the limit \( \lim_{x \to 2} (x^2 - 4) \left( \frac{1}{x + 2} + \frac{1}{x - 2} \right) \), we will follow these steps:

### Step 1: Substitute \( x = 2 \)
First, we will substitute \( x = 2 \) into the expression to check if we get a determinate form.

\[ x^2 - 4 = 2^2 - 4 = 4 - 4 = 0 \]

Now, substituting into the second part:

\[ \frac{1}{x + 2} + \frac{1}{x - 2} = \frac{1}{2 + 2} + \frac{1}{2 - 2} = \frac{1}{4} + \frac{1}{0} \]

This results in \( \frac{1}{4} + \infty \), which gives us \( 0 \times \infty \). This is an indeterminate form.

### Step 2: Rewrite the Expression
Next, we will rewrite the expression to resolve the indeterminate form. We know that \( x^2 - 4 \) can be factored as \( (x - 2)(x + 2) \):

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

### Step 3: Find a Common Denominator
Now, we will combine the fractions in the limit:

\[ \frac{1}{x + 2} + \frac{1}{x - 2} = \frac{(x - 2) + (x + 2)}{(x + 2)(x - 2)} = \frac{2x}{(x + 2)(x - 2)} \]

### Step 4: Substitute Back into the Limit
Substituting this back into our limit gives:

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

### Step 5: Cancel Out Terms
We can cancel \( (x - 2) \) and \( (x + 2) \):

\[ \lim_{x \to 2} 2x \]

### Step 6: Evaluate the Limit
Now we can substitute \( x = 2 \):

\[ 2 \cdot 2 = 4 \]

Thus, the limit is:

\[ \boxed{4} \]