The image contains several limit problems that need to be solved. Here are the problems extracted from the image:

1. $\lim_{x \to 2} \frac{3x - 6}{x - 2}$
2. $\lim_{x \to 6} \frac{x^2 - 9}{x - 4}$
3. $\lim_{x \to 2} \frac{\sqrt{x + 14} - 4}{-2x + 4}$
4. $\lim_{x \to -1} f(x)$ where \begin{cases} -2x^2 + 4 & \text{for } x \leq -1 \\ 3x + 5 & \text{for } x > -1 \end{cases}$$

Let's go through each one step by step:

1. $\lim_{x \to 2} \frac{3x - 6}{x - 2}$

First, factor the numerator: $\frac{3x - 6}{x - 2} = \frac{3(x - 2)}{x - 2}$ Cancel out the common factor $(x - 2)$: $\lim_{x \to 2} 3 = 3$ Thus, the limit is 3.

2. $\lim_{x \to 6} \frac{x^2 - 9}{x - 4}$

Factor the numerator: $x^2 - 9 = (x - 3)(x + 3)$ So, the expression becomes: $\frac{(x - 3)(x + 3)}{x - 4}$ Since direct substitution does not result in any indeterminate form (like 0/0), we can directly substitute $x = 6$: $\frac{(6 - 3)(6 + 3)}{6 - 4} = \frac{3 \times 9}{2} = \frac{27}{2}$ Thus, the limit is $\frac{27}{2}$.

3. $\lim_{x \to 2} \frac{\sqrt{x + 14} - 4}{-2x + 4}$

Substitute $x = 2$ into the expression: $\frac{\sqrt{2 + 14} - 4}{-2(2) + 4} = \frac{\sqrt{16} - 4}{-4 + 4} = \frac{4 - 4}{0} = \frac{0}{0}$ This is an indeterminate form, so we need to simplify. Let's multiply by the conjugate of the numerator: $\frac{\sqrt{x + 14} - 4}{-2x + 4} \cdot \frac{\sqrt{x + 14} + 4}{\sqrt{x + 14} + 4} = \frac{(x + 14) - 16}{(-2x + 4)(\sqrt{x + 14} + 4)}$ Simplify the numerator: $\frac{x - 2}{(-2x + 4)(\sqrt{x + 14} + 4)}$ Now, factor the denominator: $-2x + 4 = -2(x - 2)$ The expression becomes: $\frac{x - 2}{-2(x - 2)(\sqrt{x + 14} + 4)}$ Cancel out $(x - 2)$: $\lim_{x \to 2} \frac{1}{-2(\sqrt{x + 14} + 4)} = \frac{1}{-2(4 + 4)} = \frac{1}{-16}$ Thus, the limit is $\frac{-1}{16}$.

4. $\lim_{x \to -1} f(x)$, where