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.