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To determine the correct answer, let’s analyze the function given:

\[
g(x) = \frac{x^2 - 25}{3x^2 - 27x + 60}
\]

### Step 1: Simplify the Function
1. Factor the numerator:
   \[
   x^2 - 25 = (x - 5)(x + 5)
   \]

2. Factor the denominator:
   \[
   3x^2 - 27x + 60 = 3(x^2 - 9x + 20) = 3(x - 5)(x - 4)
   \]

So the function simplifies to:
\[
g(x) = \frac{(x - 5)(x + 5)}{3(x - 5)(x - 4)}
\]

### Step 2: Cancel Common Terms
The \((x - 5)\) terms cancel out, but this cancellation implies a hole at \(x = 5\). The simplified function is:
\[
g(x) = \frac{x + 5}{3(x - 4)}, \quad x \neq 5
\]

### Step 3: Evaluate the Limit as \( x \to 5 \)
To find \(\lim_{x \to 5} g(x)\), we substitute \(x = 5\) in the simplified function:
\[
\lim_{x \to 5} g(x) = \frac{5 + 5}{3(5 - 4)} = \frac{10}{3} = 3.333
\]

### Step 4: Determine the Correct Answer
Since there’s a hole at \(x = 5\), \(g(x)\) is undefined at \(x = 5\), but the limit exists and is equal to \(3.333\).

The correct answer is:
- **Option 4**: \(\lim_{x \to 5} g(x) = 3.333\), and \(g(x)\) has a hole at \(x = 5\).

Would you like a further breakdown of any steps? 

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### Related Questions:
1. How do we identify holes in a rational function?
2. What is the difference between holes and vertical asymptotes in rational functions?
3. Why is \(g(5)\) undefined even though the limit exists?
4. Can we always cancel terms like \((x - 5)\) in a rational function?
5. How can we evaluate limits in rational functions using factoring?

### Tip:
When factoring a rational function, always look for common terms in the numerator and denominator to identify possible holes or simplifications.

If g is defined as g(x) = (x^2 - 25) / (3x^2 - 27x + 60), which of the following statements is true?

Explore the rational function g(x) = (x^2 - 25) / (3x^2 - 27x + 60) to determine the correct statement about its behavior at x = 5, including concepts of limits, holes, and vertical asymptotes. Suitable for high school students in Grades 11-12.

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