Let's go through the questions from the image one by one.

1. Evaluate each function:

a) $f(3)$

Given: $f(x) = \frac{x^2 - 9}{x - 3}$

Factor the numerator: $f(x) = \frac{(x - 3)(x + 3)}{x - 3}$ For $x = 3$, the denominator becomes zero, so the function is undefined. Hence: $f(3) = \text{DNE (Does Not Exist)}$

b) $g(3)$

Given: $g(x) = \frac{x^2 + 6x}{x^2 - 3x}$ Factor both the numerator and denominator: $g(x) = \frac{x(x + 6)}{x(x - 3)}$ At $x = 3$, the denominator is zero, so $g(3) = \text{DNE}$.

c) $h(3)$

Given: $h(x) = \frac{x^2 - 8x - 33}{x^2 + 4x + 3}$ Substitute $x = 3$: $h(3) = \frac{3^2 - 8(3) - 33}{3^2 + 4(3) + 3} = \frac{9 - 24 - 33}{9 + 12 + 3} = \frac{-48}{24} = -2$ So: $h(3) = -2$

2. Evaluate each limit:

a) $\lim_{x \to 3} f(x)$

Since: $f(x) = \frac{(x - 3)(x + 3)}{x - 3}$ Cancel the $(x - 3)$ terms: $\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 6$

b) $\lim_{x \to 3} g(x)$

For $g(x) = \frac{x(x + 6)}{x(x - 3)}$, simplify: $g(x) = \frac{x + 6}{x - 3}$ As $x \to 3$, the denominator tends to zero, so: $\lim_{x \to 3} g(x) = \text{DNE (Undefined)}$

c) $\lim_{x \to 3} h(x)$

Since $h(x)$ is defined at $x = 3$, the limit can be evaluated as the function value: $\lim_{x \to 3} h(x) = h(3) = -2$

3. Continuity at $x = 3$:

• $f(x)$ is not continuous at $x = 3$ because $f(3)$ does not exist.
• $g(x)$ is not continuous at $x = 3$ because $g(3)$ does not exist.
• $h(x)$ is continuous at $x = 3$ because both $h(3)$ exists and ( \lim_{x