To solve the limit problem given in the image, we have the following expression:

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

Step 1: Factor the denominator

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

The quadratic equation is in the form $ax^2 + bx + c$. We need to find two numbers that multiply to $-63$ (the constant term) and add to $-2$ (the coefficient of $x$).

The numbers $-9$ and $7$ satisfy this condition:

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

Step 2: Simplify the expression

The limit now becomes:

$\lim_{{x \to 7}} \frac{x^2 + 7x}{(x - 9)(x + 7)}$

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

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

So the limit simplifies to:

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

Now, 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:

$\frac{7}{7 - 9} = \frac{7}{-2} = -\frac{7}{2}$

Final Answer

The limit is:

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

Would you like more details or have any questions?

Here are some related questions you might find helpful:

1. How do you factor quadratic expressions?
2. What are common limit laws and how do they apply?
3. What is the significance of canceling out terms in a limit problem?
4. How do you handle indeterminate forms in limits?
5. What is L'Hôpital's rule and when can it be applied?

Tip: When solving limits, always check for common factors that can simplify the expression, especially when the direct substitution results in an indeterminate form like $\frac{0}{0}$.