Given that

 lim *x*→2 *f*(*x*) = 9    lim *x*→2 *g*(*x*) = −2    lim *x*→2 *h*(*x*) = 0,

find the limits, if they exist. (If an answer does not exist, enter DNE.)

(a)    

lim *x*→2 [*f*(*x*) + 5*g*(*x*)]




(b)    

lim *x*→2 [*g*(*x*)]3




(c)    

lim *x*→2 

*f*(*x*)




(d)    

lim *x*→2 

4*f*(*x*)

*g*(*x*)




(e)    

lim *x*→2 

*g*(*x*)

*h*(*x*)




(f)    

lim *x*→2 

*g*(*x*)*h*(*x*)

*f*(*x*)


Let's solve each limit using the given information:

- \(\lim_{x \to 2} f(x) = 9\)
- \(\lim_{x \to 2} g(x) = -2\)
- \(\lim_{x \to 2} h(x) = 0\)

We will apply the rules of limits, such as the sum, product, and power rules, and analyze each part:

---

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

By the sum rule for limits, we can split the limit into two parts:

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

Using the given limits:

\[
\lim_{x \to 2} f(x) = 9, \quad \lim_{x \to 2} g(x) = -2
\]

\[
9 + 5 \cdot (-2) = 9 - 10 = -1
\]

So, the answer is \(-1\).

---

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

By the power rule for limits, we raise the limit of \(g(x)\) to the power 3:

\[
\lim_{x \to 2} [g(x)]^3 = [\lim_{x \to 2} g(x)]^3 = (-2)^3 = -8
\]

So, the answer is \(-8\).

---

### (c) \(\lim_{x \to 2} f(x)\)

This is given directly in the problem:

\[
\lim_{x \to 2} f(x) = 9
\]

So, the answer is \(9\).

---

### (d) \(\lim_{x \to 2} 4f(x)g(x)\)

By the product rule for limits:

\[
\lim_{x \to 2} 4f(x)g(x) = 4 \cdot \lim_{x \to 2} f(x) \cdot \lim_{x \to 2} g(x)
\]

Using the given limits:

\[
4 \cdot 9 \cdot (-2) = 4 \cdot (-18) = -72
\]

So, the answer is \(-72\).

---

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

By the product rule for limits:

\[
\lim_{x \to 2} g(x) h(x) = \lim_{x \to 2} g(x) \cdot \lim_{x \to 2} h(x)
\]

Using the given limits:

\[
-2 \cdot 0 = 0
\]

So, the answer is \(0\).

---

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

By the quotient rule for limits (if the denominator's limit is non-zero):

\[
\lim_{x \to 2} \frac{g(x)h(x)}{f(x)} = \frac{\lim_{x \to 2} g(x) h(x)}{\lim_{x \to 2} f(x)}
\]

Using the given limits:

\[
\frac{-2 \cdot 0}{9} = \frac{0}{9} = 0
\]

So, the answer is \(0\).

---

### Final Answers:

(a) \(-1\)

(b) \(-8\)

(c) \(9\)

(d) \(-72\)

(e) \(0\)

(f) \(0\)

---

Would you like more details on any of these steps, or any clarifications? Here are some related questions to consider:

1. How does the product rule for limits work?
2. What are the conditions under which the limit of a quotient exists?
3. Can limits always be split into sums and products? Why?
4. How do you handle limits when one part tends to zero?
5. What are the key differences between one-sided and two-sided limits?

**Tip:** Remember, when dealing with limits of products or quotients, always check if any part approaches zero to avoid undefined behavior!

Given that lim x→2 f(x) = 9, lim x→2 g(x) = −2, lim x→2 h(x) = 0, find the limits, if they exist: (a) lim x→2 [f(x) + 5g(x)], (b) lim x→2 [g(x)]^3, (c) lim x→2 f(x), (d) lim x→2 4f(x)g(x), (e) lim x→2 g(x)h(x), (f) lim x→2 g(x)h(x)/f(x)

Solve limit problems involving sum, product, and quotient of functions at x=2 using given limit values for f(x), g(x), and h(x). Learn how to apply the sum, product, and quotient rules for limits. Ideal for students in advanced high school calculus or undergraduate math courses.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation