Let's solve both problems step-by-step.

Problem 9

$\lim_{{x \to -1}} \frac{x^2 + 2}{x + 3}$

1. Substitute $x = -1$: $\frac{(-1)^2 + 2}{-1 + 3} = \frac{1 + 2}{2} = \frac{3}{2}$

So, the answer is d. $\frac{3}{2}$.

Problem 10

$\lim_{{x \to 4}} \frac{x^2 - 16}{x - 4}$

1. Factor the numerator $x^2 - 16$ as $(x - 4)(x + 4)$: $\frac{(x - 4)(x + 4)}{x - 4}$

2. Cancel $x - 4$ in the numerator and denominator: $= x + 4$

3. Substitute $x = 4$: $4 + 4 = 8$

So, the answer is d. 8.

Would you like a detailed explanation on any specific step or have any further questions?

Related Questions

1. How do you evaluate a limit when direct substitution results in $\frac{0}{0}$?
2. What is L'Hôpital's Rule and when can it be applied to find limits?
3. How can factoring help in solving limits involving polynomials?
4. What are common strategies for limits involving rational functions?
5. How can limits be used to find derivatives?

Tip

When facing indeterminate forms like $\frac{0}{0}$ in limits, always consider factoring, simplifying, or using L'Hôpital's Rule.