We are given the following limits:

$\lim_{x \to 3} f(x) = 4, \quad \lim_{x \to 3} g(x) = 0, \quad \lim_{x \to 3} h(x) = 0, \quad \lim_{x \to 3} \frac{g(x)}{h(x)} = 3$

We are asked to find the limit:

$\lim_{x \to 3} \frac{3(g(x)f(x)) + 2(g(x))^2}{h(x)}$

Step-by-step solution:

1. Simplify the expression:

We start by simplifying the numerator:

$3(g(x)f(x)) + 2(g(x))^2$

Using the limits provided, we can substitute $\lim_{x \to 3} f(x) = 4$ and $\lim_{x \to 3} g(x) = 0$:

• $3(g(x)f(x)) = 3 \times g(x) \times f(x) \to 3 \times 0 \times 4 = 0$
• $2(g(x))^2 = 2 \times (g(x))^2 \to 2 \times 0^2 = 0$

Thus, the numerator approaches:

$0 + 0 = 0$

2. Apply the limit to the denominator:

We are given:

$\lim_{x \to 3} h(x) = 0$

So the expression becomes a $\frac{0}{0}$ indeterminate form. To resolve this, we use the additional information:

$\lim_{x \to 3} \frac{g(x)}{h(x)} = 3$

3. Rewrite the expression using this limit:

We can factor the numerator as follows:

$\frac{3(g(x)f(x)) + 2(g(x))^2}{h(x)} = \frac{g(x)}{h(x)} \times \left(3f(x) + 2g(x)\right)$

Now, substitute the limits into this expression:

• $\lim_{x \to 3} \frac{g(x)}{h(x)} = 3$
• $\lim_{x \to 3} f(x) = 4$
• $\lim_{x \to 3} g(x) = 0$

Thus, the expression simplifies to:

$3 \times (3 \times 4 + 2 \times 0) = 3 \times 12 = 36$

Final Answer:

$\lim_{x \to 3} \frac{3(g(x)f(x)) + 2(g(x))^2}{h(x)} = 36$

Would you like more details or explanations?

Here are some related questions:

1. What are indeterminate forms in limits and how do we resolve them?
2. How does factoring help in simplifying complex limit expressions?
3. What is the role of L'Hôpital's Rule in resolving limits of indeterminate forms?
4. How can limits help in understanding the behavior of functions near critical points?
5. Why is the limit $\lim_{x \to 3} \frac{g(x)}{h(x)} = 3$ crucial in solving this problem?

Tip: Always check for indeterminate forms before applying limit rules to avoid incorrect conclusions.