Let's go through each of the given limit problems one by one using the provided information.

Given: $\lim_{x \to 2} f(x) = 4, \quad \lim_{x \to 2} g(x) = -2, \quad \lim_{x \to 2} h(x) = 0$ We'll compute the limits as follows:

(a) $\lim_{x \to 2} [f(x) + 5g(x)]$

We can split this limit using the sum rule of limits:

$\lim_{x \to 2} [f(x) + 5g(x)] = \lim_{x \to 2} f(x) + 5 \cdot \lim_{x \to 2} g(x)$

Substitute the given values:

$= 4 + 5(-2) = 4 - 10 = -6$

So, the limit is $-6$.

(b) $\lim_{x \to 2} [g(x)]^3$

By the power rule of limits:

$\lim_{x \to 2} [g(x)]^3 = \left(\lim_{x \to 2} g(x)\right)^3$

Substitute the given value for $g(x)$:

$= (-2)^3 = -8$

So, the limit is $-8$.

(c) $\lim_{x \to 2} \sqrt{f(x)}$

By the square root rule of limits:

$\lim_{x \to 2} \sqrt{f(x)} = \sqrt{\lim_{x \to 2} f(x)}$

Substitute the given value for $f(x)$:

$= \sqrt{4} = 2$

So, the limit is 2.

(d) $\lim_{x \to 2} \frac{3f(x)}{g(x)}$

Using the quotient rule of limits:

$\lim_{x \to 2} \frac{3f(x)}{g(x)} = \frac{3 \cdot \lim_{x \to 2} f(x)}{\lim_{x \to 2} g(x)}$

Substitute the given values for $f(x)$ and $g(x)$:

$= \frac{3 \cdot 4}{-2} = \frac{12}{-2} = -6$

So, the limit is $-6$.

(e) $\lim_{x \to 2} \frac{g(x)}{h(x)}$

Using the quotient rule of limits:

$\lim_{x \to 2} \frac{g(x)}{h(x)} = \frac{\lim_{x \to 2} g(x)}{\lim_{x \to 2} h(x)}$

Substitute the given values:

$= \frac{-2}{0}$

This expression results in a division by zero, which means the limit does not exist. More specifically, the limit approaches $-\infty$ or $\infty$ depending on the direction of approach, but it is undefined overall.

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

Using the quotient and product rules of limits:

$\lim_{x \to 2} \frac{g(x) h(x)}{f(x)} = \frac{\left(\lim_{x \to 2} g(x)\right) \cdot \left(\lim_{x \to 2} h(x)\right)}{\lim_{x \to 2} f(x)}$

Substitute the given values for $g(x)$, $h(x)$, and $f(x)$:

$= \frac{(-2)(0)}{4} = \frac{0}{4} = 0$

So, the limit is 0.

Summary of Results:

• (a) $-6$
• (b) $-8$
• (c) $2$
• (d) $-6$
• (e) Does not exist (division by zero)
• (f) $0$

Would you like any additional details or further explanations on any part?

Here are 5 related questions for practice:

1. What happens if $\lim_{x \to 2} g(x) = 0$ in problem (d)?
2. How would you solve $\lim_{x \to a} f(x)g(x)$ if $\lim_{x \to a} g(x) = 0$?
3. Can you explain the difference between removable and non-removable discontinuities?
4. How does the limit definition change if $g(x) \to \infty$ as $x \to 2$?
5. What are the conditions for a limit to exist when dealing with quotient functions?

Tip: Always check for the possibility of dividing by zero when dealing with limit problems involving quotient functions.