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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.

Suppose that f and g are non-constant functions, defined for all x near 0, that satisfy: lim(x→0) f(x) = 3 and lim(x→0) g(x) = 2. Evaluate the limit lim(x→0) (g(x)f(x) - 2f(x)) / ([g(x)]² - 4).

Learn how to evaluate a limit involving indeterminate forms using L'Hopital's Rule. The problem involves calculating lim(x→0) (g(x)f(x) - 2f(x)) / ([g(x)]² - 4) where g(x) and f(x) are functions near x = 0. Suitable for students studying calculus.

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