We are given that the functions $f(x)$ and $g(x)$ are non-constant and that:

$\lim_{x \to 0} f(x) = 3 \quad \text{and} \quad \lim_{x \to 0} g(x) = 2.$

We are asked to evaluate the following limit:

$\lim_{x \to 0} \frac{g(x) f(x) - 2 f(x)}{[g(x)]^2 - 4}.$

Step 1: Evaluate the numerator and denominator limits separately

First, let's break this down into components. We can evaluate the limits of the numerator and the denominator separately.

• Numerator:

$g(x) f(x) - 2 f(x)$ Using the given limits: [ \lim_{x \to 0} \left[ g(x) f(x) - 2 f(x) \right] = \lim_{x \to 0} \left[ 2 \cdot 3 - 2 \cdot 3 \right] = 6 - 6 = 0. ]

• Denominator:

$[g(x)]^2 - 4$ Again using the given limit: [ \lim_{x \to 0} \left[ [g(x)]^2 - 4 \right] = \lim_{x \to 0} \left[ 2^2 - 4 \right] = 4 - 4 = 0. ]

Both the numerator and denominator approach zero as $x \to 0$, indicating that we have a $\frac{0}{0}$ indeterminate form. This suggests we can apply L'Hopital's Rule to resolve the limit.

Step 2: Apply L'Hopital's Rule

To apply L'Hopital's Rule, we take the derivatives of the numerator and denominator.

• Derivative of the numerator $g(x) f(x) - 2 f(x)$: Using the product rule: $\frac{d}{dx} \left[ g(x) f(x) - 2 f(x) \right] = g'(x) f(x) + g(x) f'(x) - 2 f'(x).$

• Derivative of the denominator $[g(x)]^2 - 4$: Using the chain rule: [ \frac{d}{dx} \left[ [g(x)]^2 - 4 \right] = 2 g(x) g'(x). ]

Step 3: Evaluate the limit of the derivatives

Now, evaluate the new limit using these derivatives.

$\lim_{x \to 0} \frac{g'(x) f(x) + g(x) f'(x) - 2 f'(x)}{2 g(x) g'(x)}.$

Substitute the known values $\lim_{x \to 0} f(x) = 3$ and $\lim_{x \to 0} g(x) = 2$:

$\lim_{x \to 0} \frac{g'(0) \cdot 3 + 2 \cdot f'(0) - 2 f'(0)}{2 \cdot 2 \cdot g'(0)} = \lim_{x \to 0} \frac{3 g'(0)}{4 g'(0)}.$

If $g'(0) \neq 0$, the terms simplify:

$\lim_{x \to 0} \frac{3}{4}.$

Thus, the value of the original limit is $\frac{3}{4}$.

Would you like any additional details or have any further questions? Here are some related questions:

1. How does L'Hopital's Rule work for other indeterminate forms?
2. What are some situations where L'Hopital's Rule might fail?
3. Can we solve this problem using a Taylor series instead of L'Hopital's Rule?
4. What is the geometric interpretation of limits like this?
5. How would the solution change if the limits of $f(x)$ and $g(x)$ were different?

Tip: Always check if a limit results in an indeterminate form before applying L'Hopital's Rule. This ensures that the rule is applicable.