Math Problem Statement

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) + 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)

Solution

Let's solve each limit using the given information:

  • limx2f(x)=9\lim_{x \to 2} f(x) = 9
  • limx2g(x)=2\lim_{x \to 2} g(x) = -2
  • limx2h(x)=0\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) limx2[f(x)+5g(x)]\lim_{x \to 2} [f(x) + 5g(x)]

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

limx2[f(x)+5g(x)]=limx2f(x)+5limx2g(x)\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:

limx2f(x)=9,limx2g(x)=2\lim_{x \to 2} f(x) = 9, \quad \lim_{x \to 2} g(x) = -2

9+5(2)=910=19 + 5 \cdot (-2) = 9 - 10 = -1

So, the answer is 1-1.


(b) limx2[g(x)]3\lim_{x \to 2} [g(x)]^3

By the power rule for limits, we raise the limit of g(x)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 8-8.


(c) limx2f(x)\lim_{x \to 2} f(x)

This is given directly in the problem:

limx2f(x)=9\lim_{x \to 2} f(x) = 9

So, the answer is 99.


(d) limx24f(x)g(x)\lim_{x \to 2} 4f(x)g(x)

By the product rule for limits:

limx24f(x)g(x)=4limx2f(x)limx2g(x)\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:

49(2)=4(18)=724 \cdot 9 \cdot (-2) = 4 \cdot (-18) = -72

So, the answer is 72-72.


(e) limx2g(x)h(x)\lim_{x \to 2} g(x) h(x)

By the product rule for limits:

limx2g(x)h(x)=limx2g(x)limx2h(x)\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:

20=0-2 \cdot 0 = 0

So, the answer is 00.


(f) limx2g(x)h(x)f(x)\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):

limx2g(x)h(x)f(x)=limx2g(x)h(x)limx2f(x)\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:

209=09=0\frac{-2 \cdot 0}{9} = \frac{0}{9} = 0

So, the answer is 00.


Final Answers:

(a) 1-1

(b) 8-8

(c) 99

(d) 72-72

(e) 00

(f) 00


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!

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Math Problem Analysis

Mathematical Concepts

Limits
Sum Rule for Limits
Product Rule for Limits
Quotient Rule for Limits

Formulas

lim x→c [f(x) + g(x)] = lim x→c f(x) + lim x→c g(x)
lim x→c [f(x) * g(x)] = lim x→c f(x) * lim x→c g(x)
lim x→c [f(x)/g(x)] = lim x→c f(x) / lim x→c g(x) (if lim x→c g(x) ≠ 0)

Theorems

Sum Rule for Limits
Product Rule for Limits
Quotient Rule for Limits

Suitable Grade Level

Undergraduate Mathematics or Advanced High School Calculus