We are tasked with finding the limit of the following expression as $x \to -1$:

$\lim_{{x \to -1}} \frac{-2x - 2}{x^2 - 6x - 7}$

Step 1: Evaluate the function at $x = -1$

First, substitute $x = -1$ into the expression:

$\text{Numerator: } -2(-1) - 2 = 2 - 2 = 0$ $\text{Denominator: } (-1)^2 - 6(-1) - 7 = 1 + 6 - 7 = 0$

Since both the numerator and denominator equal zero, we have an indeterminate form $\frac{0}{0}$. Therefore, we need to simplify the expression further, usually by factoring.

Step 2: Factor the numerator and denominator

• The numerator $-2x - 2$ can be factored as: $-2x - 2 = -2(x + 1)$
• The denominator $x^2 - 6x - 7$ can be factored by finding two numbers that multiply to $-7$ and add to $-6$. These numbers are $-7$ and $1$, so: $x^2 - 6x - 7 = (x - 7)(x + 1)$

Step 3: Simplify the expression

Now substitute the factored forms back into the limit: $\lim_{{x \to -1}} \frac{-2(x + 1)}{(x - 7)(x + 1)}$

The common factor $(x + 1)$ cancels out: $\lim_{{x \to -1}} \frac{-2}{x - 7}$

Step 4: Evaluate the simplified limit

Now, substitute $x = -1$ into the simplified expression: $\frac{-2}{-1 - 7} = \frac{-2}{-8} = \frac{1}{4}$

Final Answer:

$\lim_{{x \to -1}} \frac{-2x - 2}{x^2 - 6x - 7} = \frac{1}{4}$

Would you like further details or have any questions?

Here are five related questions you might find helpful:

1. How do you determine if a limit results in an indeterminate form?
2. What are other common methods for resolving indeterminate forms besides factoring?
3. How do you find the limit as $x \to \infty$?
4. Can L'Hopital's Rule be applied here? If so, how would it work?
5. How do you factor a quadratic equation efficiently?

Tip: Always check for common factors in both the numerator and denominator to simplify limits involving indeterminate forms.