1. Chapter 3 Limits and the Derivative
Section 2
Infinite Limits and Limits at Infinity
(Part 1)

2. Objectives for Section 3.2 Infinite Limits and Limits at Infinity
The student will understand the concept of infinite limits.
The student will be able to calculate limits at infinity.
Barnett/Ziegler/Byleen Business Calculus 12e

3. Example 1 Recall from the first lesson: lim 𝑥→ 0 − 1 𝑥 =
𝐷𝑁𝐸 −∞ ∞
Barnett/Ziegler/Byleen Business Calculus 12e

4. Infinite Limits and Vertical Asymptotes
Definition:
If the graph of y = f (x) has a vertical asymptote of x = a, then as x approaches a from the left or right, then f(x) approaches either  or -.
Vertical asymptotes (and holes) are called points of discontinuity.
Barnett/Ziegler/Byleen Business Calculus 12e

5. Example 2
Let
Identify all holes and asymptotes and find the left and right hand limits as x approaches the vertical asymptotes.
𝑓 𝑥 = (𝑥+2)(𝑥−1) (𝑥+1)(𝑥−1)
= 𝑥+2 𝑥+1
𝐻𝑜𝑙𝑒:(1, 1.5)
𝑉𝐴:𝑥=−1
𝐻𝐴:𝑦=1
Barnett/Ziegler/Byleen Business Calculus 12e

6. Example 2 (continued) Vertical Asymptote Hole Horizontal Asymptote
lim 𝑥→ −1 − 𝑥+2 𝑥+1 =
lim 𝑥→ − 𝑥+2 𝑥+1 =
lim 𝑥→−1 𝑥+2 𝑥+1 = −∞ ∞
𝐷𝑁𝐸
Barnett/Ziegler/Byleen Business Calculus 12e

7. Example 3
𝑓 𝑥 = 1 (𝑥−2) 2
Let
Identify all holes and asymptotes and find the left and right hand limits as x approaches the vertical asymptotes.
𝑓 x = 1 (𝑥−2) 2
𝑁𝑜 𝐻𝑜𝑙𝑒𝑠
𝑉𝐴:𝑥=2
𝐻𝐴:𝑦=0
Barnett/Ziegler/Byleen Business Calculus 12e

8. Example 3 (continued) lim 𝑥→ 2 − 1 (𝑥−2) 2 = lim 𝑥→ 2 + 1 (𝑥−2) 2 =∞ ∞ ∞
Barnett/Ziegler/Byleen Business Calculus 12e

9. Limits at Infinity We will now study limits as x  ±.
This is the same concept as the end behavior of a graph.
Barnett/Ziegler/Byleen Business Calculus 12e

10. End Behavior Review Even degree Negative leading coefficient
Odd degree
Positive
leading
coefficient
Odd degree
Negative
leading
coefficient
Even degree
Positive
leading
coefficient
Barnett/Ziegler/Byleen Business Calculus 12e

11. Polynomial Functions Ex 4: Evaluate each limit. lim 𝑥→−∞ 𝑥 2 = ∞
Barnett/Ziegler/Byleen Business Calculus 12e

12. Rational Functions
If a rational function has a horizontal asymptote, then it determines the end behavior of the graph.
If f(x) is a rational function, then
lim 𝑥→±∞ 𝑓 𝑥 =ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒 𝑣𝑎𝑙𝑢𝑒
Barnett/Ziegler/Byleen Business Calculus 12e

13. Because the degree of the numerator < degree of the denominator.
Rational Functions
Ex 5: Evaluate
lim 𝑥→±∞ 𝑓 𝑥
𝑓 𝑥 = 5 𝑥+2
Because the degree of the numerator < degree of the denominator.
𝐻𝐴:𝑦=0
lim 𝑥→∞ 𝑓 𝑥 =
lim 𝑥→−∞ 𝑓 𝑥 =
Barnett/Ziegler/Byleen Business Calculus 12e

14. Because the degree of the numerator = degree of the denominator.
Rational Functions
Ex 6: Evaluate
lim 𝑥→±∞ 𝑓 𝑥
𝐻𝐴:𝑦= 3 2
Because the degree of the numerator = degree of the denominator.
3 2
lim 𝑥→∞ 𝑓 𝑥 =
3 2
lim 𝑥→−∞ 𝑓 𝑥 =
Barnett/Ziegler/Byleen Business Calculus 12e

15. Rational Functions
If a rational function doesn’t have a horizontal asymptote, then to determine its end behavior, take the limit of the ratio of the leading terms of the top and bottom.
Barnett/Ziegler/Byleen Business Calculus 12e

16. Because the degree of the numerator > degree of the denominator.
Rational Functions
Ex 7: Evaluate
lim 𝑥→±∞ 𝑓 𝑥
𝑓 𝑥 = 2 𝑥 5 − 𝑥 3 −1 6 𝑥 3 +2 𝑥 2 −7
Because the degree of the numerator > degree of the denominator.
𝐻𝐴:𝑁𝑜𝑛𝑒
lim 𝑥→∞ 2 𝑥 5 6 𝑥 3 =
lim 𝑥→∞ 𝑥 2 3 = ∞
lim 𝑥→−∞ 2 𝑥 5 6 𝑥 3 =
lim 𝑥→−∞ 𝑥 2 3 = ∞
Barnett/Ziegler/Byleen Business Calculus 12e

17. Rational Functions Ex 8: Evalaute lim 𝑥→±∞ 𝑓 𝑥
𝑓 𝑥 = 5 𝑥 6 +3𝑥 2 𝑥 5 −𝑥−5
𝐻𝐴:𝑁𝑜𝑛𝑒
lim 𝑥→∞ 5 𝑥 6 2 𝑥 5 =
lim 𝑥→∞ 5𝑥 2 = ∞
lim 𝑥→−∞ 5 𝑥 6 2 𝑥 5 =
lim 𝑥→−∞ 5𝑥 2 = −∞
Barnett/Ziegler/Byleen Business Calculus 12e

18. Homework
#3-2A: Pg 150
(3-15 mult. of 3,
17, 19, odd,
39, 43, 45, 61, 65)
Barnett/Ziegler/Byleen Business Calculus 12e

19. Chapter 3 Limits and the Derivative
Section 2
Infinite Limits and Limits at Infinity
(Part 2)

20. Objectives for Section 3.2 Infinite Limits and Limits at Infinity
The student will be able to solve applications involving limits.
2 ∞ & >
Barnett/Ziegler/Byleen Business Calculus 12e

21. Application: Business
T & C surf company makes surfboards with fixed costs at $300 per day. One day, they made 20 boards and total costs were $5100.
Assuming the total cost per day is linearly related to the number of boards made per day, write an equation for the cost function.
Write the equation for the average cost function.
Graph the average cost function: 𝑥: 1, 30 𝑦:[0, 500]
What does the average cost per board approach as production increases?
Barnett/Ziegler/Byleen Business Calculus 12e

22. Application: Business
T & C surf company makes surfboards with fixed costs at $300 per day. One day, they made 20 boards and total costs were $5100.
Assuming the total cost per day is linearly related to the number of boards made per day, write an equation for the cost function.
𝑦=𝑚𝑥+𝑏
5100=𝑚(20)+300
𝑚=240
𝑇ℎ𝑒 𝑐𝑜𝑠𝑡 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠: 𝐶 𝑥 =240𝑥+300
Barnett/Ziegler/Byleen Business Calculus 12e

23. Application: Business
T & C surf company makes surfboards with fixed costs at $300 per day. One day, they made 20 boards and total costs were $5100.
Write the equation for the average cost function.
𝐶 𝑥 = 𝐶(𝑥) 𝑥
𝐶 𝑥 = 240𝑥+300 𝑥
Barnett/Ziegler/Byleen Business Calculus 12e

24. Application: Business
T & C surf company makes surfboards with fixed costs at $300 per day. One day, they made 20 boards and total costs were $5100.
Graph the average cost function: 𝑥: 1, 30 𝑦:[0, 500]
Number of surfboards
Average cost per day
Barnett/Ziegler/Byleen Business Calculus 12e

25. Application: Business
T & C surf company makes surfboards with fixed costs at $300 per day. One day, they made 20 boards and total costs were $5100.
What does the average cost per board approach as production increases?
Number of surfboards
Average cost per day
𝐶 𝑥 = 240𝑥+300 𝑥
As the number of boards increases, the average cost approaches $240 per board.
Barnett/Ziegler/Byleen Business Calculus 12e

26. Application: Medicine
A drug is administered to a patient through an IV drip. The drug concentration (mg per milliliter) in the patient’s bloodstream t hours after the drip was started is modeled by the equation:
What is the drug concentration after 2 hours?
Evaluate and interpret the meaning of the limit:
𝐶 𝑡 = 5𝑡 𝑡+50 𝑡
lim 𝑡→∞ 𝐶(𝑡)
Barnett/Ziegler/Byleen Business Calculus 12e

27. Application: Medicine
A drug is administered to a patient through an IV drip. The drug concentration (mg per milliliter) in the patient’s bloodstream t hours after the drip was started is modeled by the equation:
What is the drug concentration after 2 hours?
𝐶 𝑡 = 5𝑡 𝑡+50 𝑡
𝐶 2 = 5(2) ≈4.8
After 2 hours, the concentration of the drug is 4.8 mg/ml.
Barnett/Ziegler/Byleen Business Calculus 12e

28. Application: Medicine
A drug is administered to a patient through an IV drip. The drug concentration (mg per milliliter) in the patient’s bloodstream t hours after the drip was started is modeled by the equation:
Evaluate and interpret the meaning of the limit:
𝐶 𝑡 = 5𝑡 𝑡+50 𝑡
lim 𝑡→∞ 𝐶(𝑡)
lim 𝑡→∞ 5𝑡 𝑡+50 𝑡 =0
As time passes, the drug concentration approaches 0 mg/ml.
Barnett/Ziegler/Byleen Business Calculus 12e

29. Homework
#3-2B: Pg 150
(2-8 even, 11, 13, 18,
34, 36, 37, 41, 49,
63, 67, 69, 73, 76)
Barnett/Ziegler/Byleen Business Calculus 12e