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### 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:
\[ h(3) = -2 \]

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### 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:
\[ \lim_{x \to 3} h(x) = h(3) = -2 \]

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### 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

Evaluate the functions f(x), g(x), h(x) and their limits as x approaches 3. Check for continuity at x = 3 for each function.

This problem explores limits and continuity of functions f(x), g(x), and h(x) as x approaches 3. Learn how to evaluate limits, simplify rational expressions, and determine continuity at a point. Suitable for students in Grades 10-12.

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