We are tasked with finding the following limit:

$\lim_{x \to -1} \frac{x + 1}{x^2 - 4x - 5}$

Step 1: Substitute $x = -1$ into the expression.

First, we substitute $x = -1$ into the numerator and denominator:

Numerator:

$x + 1 = (-1) + 1 = 0$

Denominator:

$x^2 - 4x - 5 = (-1)^2 - 4(-1) - 5 = 1 + 4 - 5 = 0$

So, directly substituting gives us a form $\frac{0}{0}$, which is indeterminate. This indicates that we need to factor and simplify the expression.

Step 2: Factor the denominator

The denominator is a quadratic expression. Let's factor it:

$x^2 - 4x - 5 = (x - 5)(x + 1)$

Step 3: Simplify the expression

Now, rewrite the limit with the factored denominator:

$\lim_{x \to -1} \frac{x + 1}{(x - 5)(x + 1)}$

Notice that $x + 1$ appears in both the numerator and denominator, so we can cancel it:

$\lim_{x \to -1} \frac{1}{x - 5}$

Step 4: Substitute $x = -1$

Now, substitute $x = -1$ into the simplified expression:

$\frac{1}{-1 - 5} = \frac{1}{-6}$

Final Answer:

$\lim_{x \to -1} \frac{x + 1}{x^2 - 4x - 5} = -\frac{1}{6}$

Would you like more details or have any further questions?

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Here are 5 related questions for further practice:

1. What is the limit $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$?
2. How would you solve $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$?
3. Can you factor and simplify $x^2 - 5x + 6$ and find the limit as $x \to 2$?
4. What is the behavior of $\lim_{x \to -2} \frac{x^2 + 2x}{x + 2}$?
5. How do we handle limits that result in indeterminate forms like $\frac{0}{0}$?

Tip: Always factor expressions when you encounter the indeterminate form $\frac{0}{0}$, as it often simplifies the limit.