Let's evaluate each limit step by step.

i. $\lim_{x \to -4} \frac{x^2 + x - 12}{x^2 - 16}$

1. Factorize both the numerator and the denominator: $x^2 + x - 12 = (x - 3)(x + 4)$ $x^2 - 16 = (x - 4)(x + 4)$ So, the expression becomes: $\lim_{x \to -4} \frac{(x - 3)(x + 4)}{(x - 4)(x + 4)}$

2. Simplify: Cancel out $(x + 4)$ from both numerator and denominator: $\lim_{x \to -4} \frac{x - 3}{x - 4}$

3. Substitute $x = -4$: $\frac{-4 - 3}{-4 - 4} = \frac{-7}{-8} = \frac{7}{8}$

So, $\lim_{x \to -4} \frac{x^2 + x - 12}{x^2 - 16} = \frac{7}{8}.$

ii. $\lim_{x \to -2} \frac{x^3 + 2x^2 - x - 2}{x^2 - 4}$

1. Factorize the denominator: $x^2 - 4 = (x - 2)(x + 2)$

2. Factorize the numerator: Using synthetic division or factoring techniques, we find: $x^3 + 2x^2 - x - 2 = (x + 2)(x^2 + x - 1)$

3. Simplify: $\lim_{x \to -2} \frac{(x + 2)(x^2 + x - 1)}{(x - 2)(x + 2)}$ Cancel $(x + 2)$ from both numerator and denominator: $\lim_{x \to -2} \frac{x^2 + x - 1}{x - 2}$

4. Substitute $x = -2$: $\frac{(-2)^2 + (-2) - 1}{-2 - 2} = \frac{4 - 2 - 1}{-4} = \frac{1}{-4} = -\frac{1}{4}$

So, $\lim_{x \to -2} \frac{x^3 + 2x^2 - x - 2}{x^2 - 4} = -\frac{1}{4}.$

iii. $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

This is an indeterminate form $\frac{0}{0}$, so we apply L'Hopital's Rule (differentiating numerator and denominator):

1. Differentiate numerator and denominator:

   • Derivative of the numerator: $\frac{d}{dx} (\sqrt{x} - 2) = \frac{1}{2\sqrt{x}}$
   • Derivative of the denominator: $\frac{d}{dx} (x - 4) = 1$

2. Apply L'Hopital's Rule: $\lim_{x \to 4} \frac{\frac{1}{2\sqrt{x}}}{1} = \frac{1}{2\sqrt{4}} = \frac{1}{2 \times 2} = \frac{1}{4}$

So, $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{4}.$

iv. $\lim_{x \to \infty} \frac{5x + 2}{7x - x^2}$

1. Divide both the numerator and the denominator by $x^2$: $\lim_{x \to \infty} \frac{\frac{5x}{x^2} + \frac{2}{x^2}}{\frac{7x}{x^2} - \frac{x^2}{x^2}} = \lim_{x \to \infty} \frac{\frac{5}{x} + \frac{2}{x^2}}{\frac{7}{x} - 1}$

2. As $x \to \infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ vanish: $\lim_{x \to \infty} \frac{0 + 0}{0 - 1} = \frac{0}{-1} = 0$

So, $\lim_{x \to \infty} \frac{5x + 2}{7x - x^2} = 0.$

Final Answers:

i. $\frac{7}{8}$

ii. $-\frac{1}{4}$

iii. $\frac{1}{4}$

iv. $0$

---

Do you need further details or explanations on any of these problems?

Here are 5 related questions for you:

1. What is L'Hopital's Rule and when is it applicable?
2. How do you factor cubic polynomials?
3. How can limits be used to find horizontal asymptotes of functions?
4. What are indeterminate forms, and how can they be resolved?
5. How do you recognize the need to apply L'Hopital's Rule in a limit problem?

Tip: Always check if a limit problem results in an indeterminate form before proceeding with L'Hopital’s Rule.