We are asked to find the limit:

$\lim_{x \to c} \left( \sqrt{ \frac{[f(x) - 5]^2}{g(x)^3}} \right)$

Given:

• $\lim_{x \to c} f(x) = 3$
• $\lim_{x \to c} g(x) = -2$

Step-by-step Solution:

1. Break down the limit: The limit expression can be broken down into smaller parts: [ \lim_{x \to c} \left( \sqrt{ \frac{[f(x) - 5]^2}{g(x)^3}} \right) = \sqrt{ \lim_{x \to c} \left( \frac{[f(x) - 5]^2}{g(x)^3} \right) } ] Since square roots and limits commute when the inside is non-negative, we can take the limit inside the square root.

2. Evaluate the numerator: [ \lim_{x \to c} [f(x) - 5]^2 = [ \lim_{x \to c} (f(x) - 5) ]^2 = [ 3 - 5 ]^2 = (-2)^2 = 4 ]

3. Evaluate the denominator: $\lim_{x \to c} g(x)^3 = \left( \lim_{x \to c} g(x) \right)^3 = (-2)^3 = -8$

4. Combine the results: Now, combine the numerator and denominator: $\lim_{x \to c} \frac{[f(x) - 5]^2}{g(x)^3} = \frac{4}{-8} = -\frac{1}{2}$

5. Apply the square root: Finally, apply the square root to the result: $\sqrt{-\frac{1}{2}}$ However, this is not a real number. The square root of a negative number is imaginary. Thus, we can express the result as: $\boxed{i \frac{1}{\sqrt{2}}}$ where $i$ is the imaginary unit.

Exact answer: $i \frac{1}{\sqrt{2}}$

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Would you like more details or have any questions?

Here are 5 follow-up questions to consider:

1. How do limits behave when involving complex numbers?
2. What are the properties of limits involving square roots?
3. How does the limit of a function change if $g(x)$ were positive instead of negative?
4. Why is the square root of a negative number expressed using $i$?
5. What happens to limits when both numerator and denominator approach zero?

Tip: Always check whether the limit produces a real or complex result when dealing with square roots of negative numbers.