Thank you for the clarification! Let's correct the solution given that $x$ is approaching $-7$.

We are given the limit:

$\lim_{{x \to -7}} \frac{x^2 + 7x}{x^2 - 2x - 63}$

Step 1: Factor the denominator

As before, let's factor the denominator $x^2 - 2x - 63$:

$x^2 - 2x - 63 = (x - 9)(x + 7)$

Step 2: Simplify the expression

The numerator $x^2 + 7x$ can also be factored:

$x^2 + 7x = x(x + 7)$

So the limit simplifies to:

$\lim_{{x \to -7}} \frac{x(x + 7)}{(x - 9)(x + 7)}$

We can cancel out the common factor $(x + 7)$ in the numerator and denominator:

$\lim_{{x \to -7}} \frac{x}{x - 9}$

Step 3: Substitute $x = -7$

Now, substitute $x = -7$ into the simplified expression:

$\frac{-7}{-7 - 9} = \frac{-7}{-16} = \frac{7}{16}$

Final Answer

The limit is:

$\lim_{{x \to -7}} \frac{x^2 + 7x}{x^2 - 2x - 63} = \frac{7}{16}$

Would you like more details or have any other questions?

Here are some related questions you might find helpful:

1. What happens when you cancel common factors in limits?
2. How do you determine when a limit exists or does not exist (DNE)?
3. What are some common pitfalls when evaluating limits?
4. How do limits apply to continuous functions?
5. What are the conditions for the existence of a limit?

Tip: Always double-check the point to which $x$ is approaching. Small changes in the value $x$ approaches can significantly affect the outcome.