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The given limit expression is:

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

Let's solve it step by step.

### Step 1: Simplify the second fraction.
The denominator \(x^2 - 3^2\) can be factored as a difference of squares:
\[
x^2 - 3^2 = (x - 3)(x + 3)
\]
Thus, the second term becomes:
\[
\frac{6}{x^2 - 3^2} = \frac{6}{(x - 3)(x + 3)}
\]

### Step 2: Combine the two fractions.
The expression becomes:
\[
\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{6}{(x - 3)(x + 3)} \right)
\]
To combine these, we need a common denominator. The common denominator will be \((x - 3)(x + 3)\). Rewriting both fractions:
\[
\frac{1}{x - 3} = \frac{x + 3}{(x - 3)(x + 3)}
\]
Thus, the expression becomes:
\[
\lim_{x \to 3} \left( \frac{x + 3}{(x - 3)(x + 3)} - \frac{6}{(x - 3)(x + 3)} \right)
\]
Now combine the numerators:
\[
\frac{(x + 3) - 6}{(x - 3)(x + 3)} = \frac{x - 3}{(x - 3)(x + 3)}
\]

### Step 3: Simplify.
Cancel out the \((x - 3)\) terms in the numerator and denominator:
\[
\frac{x - 3}{(x - 3)(x + 3)} = \frac{1}{x + 3}
\]

### Step 4: Evaluate the limit.
Now, evaluate the limit as \(x \to 3\):
\[
\lim_{x \to 3} \frac{1}{x + 3} = \frac{1}{3 + 3} = \frac{1}{6}
\]

### Final Answer:
The value of the limit is \(\frac{1}{6}\).

Would you like further clarification or details on any part of the solution?

Here are some related questions you might find helpful:
1. What is the difference of squares factorization?
2. How do you find the common denominator for rational expressions?
3. What are the basic rules for limits involving rational functions?
4. How do you handle indeterminate forms in limits?
5. How does factoring help simplify limits?

**Tip:** Always check for factorizations or common terms that can simplify complex limit expressions.

Evaluate the limit as x approaches 3 of (1/(x-3) - 6/(x^2 - 3^2)).

Step-by-step solution for calculating the limit as x approaches 3 of the expression (1/(x-3) - 6/(x^2 - 3^2)). This problem involves understanding the difference of squares factorization, rational expressions, and limit laws. Suitable for Grades 11-12 calculus students.

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