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?

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1. What is the difference of squares factorization?
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Tip: Always check for factorizations or common terms that can simplify complex limit expressions.