7. $$ \lim _{x \rightarrow 3} \frac{x^{2}-x-6}{x^{2}+x-12} $$

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To evaluate the limit:

$$
\lim _{x \rightarrow 3} \frac{x^{2}-x-6}{x^{2}+x-12}
$$

**Step 1: Substitute $$ x = 3 $$ to Check for Indeterminate Form**

$$
\text{Numerator: } 3^{2} - 3 - 6 = 9 - 3 - 6 = 0
$$
$$
\text{Denominator: } 3^{2} + 3 - 12 = 9 + 3 - 12 = 0
$$

Since both the numerator and denominator evaluate to 0, we have an indeterminate form $$ \frac{0}{0} $$. This indicates that we can factor both the numerator and the denominator to simplify the expression.

**Step 2: Factor the Numerator and Denominator**

- **Numerator ($$ x^2 - x - 6 $$):**
  We look for two numbers that multiply to $$-6$$ and add to $$-1$$. These numbers are $$-3$$ and $$2$$.
  
  $$
  x^2 - x - 6 = (x - 3)(x + 2)
  $$

- **Denominator ($$ x^2 + x - 12 $$):**
  We look for two numbers that multiply to $$-12$$ and add to $$1$$. These numbers are $$4$$ and $$-3$$.
  
  $$
  x^2 + x - 12 = (x + 4)(x - 3)
  $$

**Step 3: Simplify the Expression**

Substitute the factored forms back into the limit:

$$
\lim _{x \rightarrow 3} \frac{(x - 3)(x + 2)}{(x + 4)(x - 3)}
$$

Cancel the common factor $$(x - 3)$$:

$$
\lim _{x \rightarrow 3} \frac{x + 2}{x + 4}
$$

**Step 4: Evaluate the Simplified Limit**

Now, substitute $$ x = 3 $$ into the simplified expression:

$$
\frac{3 + 2}{3 + 4} = \frac{5}{7}
$$

**Conclusion:**

$$
\lim _{x \rightarrow 3} \frac{x^{2}-x-6}{x^{2}+x-12} = \frac{5}{7}
$$

**Answer:** The limit is 5⁄7.
The limit is $$ \frac{5}{7} $$.

7. \( \lim _{x \rightarrow 3} \frac{x^{2}-x-6}{x^{2}+x-12} \)

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