Evaluate the following limits :
`Lim_(xto3) (x^(2)-9) (1/(x+3)+1/(x-3))`

Answer

To evaluate the limit

\[ \lim_{x \to 3} (x^2 - 9) \left( \frac{1}{x + 3} + \frac{1}{x - 3} \right), \]

we will follow these steps:

### Step 1: Simplify the expression inside the limit

We start with the limit:

\[ \lim_{x \to 3} (x^2 - 9) \left( \frac{1}{x + 3} + \frac{1}{x - 3} \right). \]

First, we can find a common denominator for the fractions:

\[ \frac{1}{x + 3} + \frac{1}{x - 3} = \frac{(x - 3) + (x + 3)}{(x + 3)(x - 3)} = \frac{2x}{(x + 3)(x - 3)}. \]

### Step 2: Substitute back into the limit

Now we substitute this back into our limit:

\[ \lim_{x \to 3} (x^2 - 9) \cdot \frac{2x}{(x + 3)(x - 3)}. \]

### Step 3: Factor \(x^2 - 9\)

Notice that \(x^2 - 9\) can be factored as:

\[ x^2 - 9 = (x - 3)(x + 3). \]

So, we can rewrite the limit as:

\[ \lim_{x \to 3} (x - 3)(x + 3) \cdot \frac{2x}{(x + 3)(x - 3)}. \]

### Step 4: Cancel common terms

Now we can cancel \((x - 3)\) and \((x + 3)\) from the numerator and denominator:

\[ \lim_{x \to 3} 2x. \]

### Step 5: Evaluate the limit

Now we can directly substitute \(x = 3\):

\[ 2 \cdot 3 = 6. \]

Thus, the limit is:

\[ \lim_{x \to 3} (x^2 - 9) \left( \frac{1}{x + 3} + \frac{1}{x - 3} \right) = 6. \]

### Final Answer:
\[ \boxed{6} \]