To evaluate the limit

$\lim_{{x \to 7}} \frac{x^2 + x - 56}{x - 7}$

we first try direct substitution by plugging $x = 7$ into the expression. However, this yields:

$\frac{7^2 + 7 - 56}{7 - 7} = \frac{49 + 7 - 56}{0} = \frac{0}{0}$

This is an indeterminate form, so we need to simplify the expression. Let's factor the numerator.

1. Rewrite the numerator $x^2 + x - 56$ as a product of two binomials. We need two numbers that multiply to $-56$ and add to $1$: $x^2 + x - 56 = (x + 8)(x - 7)$

2. Substitute the factored form back into the limit expression: $\lim_{{x \to 7}} \frac{(x + 8)(x - 7)}{x - 7}$

3. Cancel out the $x - 7$ term in the numerator and denominator: $= \lim_{{x \to 7}} (x + 8)$

4. Now substitute $x = 7$ into the simplified expression: $= 7 + 8 = 15$

So, the limit is:

$\lim_{{x \to 7}} \frac{x^2 + x - 56}{x - 7} = 15$

Would you like more details or have any further questions?

Here are five related questions that expand on this topic:

1. How do we handle other types of indeterminate forms, such as $\frac{0}{0}$?
2. What are some methods for factoring complex expressions?
3. How can we use L'Hôpital's Rule to solve similar limits?
4. What is the role of simplifying before substituting in limit problems?
5. How do we approach limits involving higher degrees of indeterminate forms?

Tip: Always check for factorization or simplification opportunities in rational functions when solving limits to resolve indeterminate forms.