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The limit problem provided is:

\[
\lim_{{x \to -1}} \frac{x^2 + 6x + 5}{x^2 - 3x - 4}
\]

### Steps to Solve:

1. **Factorize the numerator and denominator**:
   - The numerator: \( x^2 + 6x + 5 \) can be factored as \( (x + 5)(x + 1) \).
   - The denominator: \( x^2 - 3x - 4 \) can be factored as \( (x - 4)(x + 1) \).

2. **Rewrite the expression**:
   \[
   \frac{x^2 + 6x + 5}{x^2 - 3x - 4} = \frac{(x + 5)(x + 1)}{(x - 4)(x + 1)}
   \]

3. **Cancel out common factors**:
   Since \( (x + 1) \) appears in both the numerator and the denominator, we can cancel it out:
   \[
   \frac{x + 5}{x - 4}, \quad x \neq -1
   \]
   (The condition \( x \neq -1 \) comes from the fact that \( x + 1 \) was canceled, and at \( x = -1 \), the original expression would have been undefined.)

4. **Substitute the limit**:
   Now, substitute \( x = -1 \) into the simplified expression:
   \[
   \frac{-1 + 5}{-1 - 4} = \frac{4}{-5} = -\frac{4}{5}
   \]

### Final Answer:
\[
\lim_{{x \to -1}} \frac{x^2 + 6x + 5}{x^2 - 3x - 4} = -\frac{4}{5}
\]

Let me know if you'd like further explanation!

### Related Questions:
1. How do you factor quadratic expressions effectively?
2. What happens when you cancel common factors in rational expressions?
3. How do you determine if a function is undefined at a certain point?
4. What are common techniques for solving limits algebraically?
5. What is the significance of continuity in determining limits?

**Tip**: Always check for common factors to simplify expressions before directly substituting values in limit problems.

lim (x→−1) (x^2 + 6x + 5) / (x^2 − 3x − 4)

Learn how to solve the limit of (x^2 + 6x + 5)/(x^2 − 3x − 4) as x approaches -1 with a detailed step-by-step breakdown. This problem covers key algebraic concepts such as factoring, simplifying rational expressions, and applying limits. Suitable for students in Grades 11-12 or early college-level calculus.

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