The problem requires evaluating the limit:

$\lim_{x \to 3} \frac{x^3 f(x) - 54}{g(x) - 1}.$

Step 1: Evaluate the numerator and denominator at $x = 3$

Using the table:

• $f(3) = 2$,
• $g(3) = 1$.

Substitute into the numerator and denominator: $x^3 f(x) - 54 = 3^3 \cdot f(3) - 54 = 27 \cdot 2 - 54 = 54 - 54 = 0,$ $g(x) - 1 = g(3) - 1 = 1 - 1 = 0.$

Since both the numerator and denominator approach $0$ as $x \to 3$, this is an indeterminate form $\frac{0}{0}$. We apply L'Hôpital's Rule.

Step 2: Differentiate the numerator and denominator

The numerator is: $x^3 f(x) - 54.$ Using the product rule for $x^3 f(x)$, its derivative is: $\frac{d}{dx}[x^3 f(x)] = 3x^2 f(x) + x^3 f'(x).$ So the derivative of the numerator is: $3x^2 f(x) + x^3 f'(x).$

The denominator is: $g(x) - 1,$ and its derivative is: $g'(x).$

Step 3: Substitute $x = 3$ into the derivatives

Using the table:

• $f(3) = 2$,
• $f'(3) = -4$,
• $g'(3) = 2$.

Substitute into the derivative of the numerator: $3x^2 f(x) + x^3 f'(x) = 3(3^2)(2) + (3^3)(-4) = 3(9)(2) + 27(-4) = 54 - 108 = -54.$

Substitute into the derivative of the denominator: $g'(x) = g'(3) = 2.$

Step 4: Compute the limit

Apply L'Hôpital's Rule: $\lim_{x \to 3} \frac{x^3 f(x) - 54}{g(x) - 1} = \frac{-54}{2} = -27.$

Final Answer:

$\boxed{-27} \quad (\text{Option C}).$

Let me know if you'd like further clarification or a deeper breakdown!

5 Related Questions:

1. What is L'Hôpital's Rule, and when can it be applied?
2. How do you differentiate products of functions like $x^3 f(x)$?
3. What happens if the derivative of the denominator is 0?
4. How do you analyze limits that approach infinity rather than 0?
5. Can this problem be solved without L'Hôpital's Rule? If so, how?

One Tip:

Always verify that both the numerator and denominator approach 0 or $\pm \infty$ before applying L'Hôpital's Rule.