Chapter 3
Limits and the
Derivative
Section 2
Infinite Limits and
Limits at Infinity
(Part 1)
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
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Example 1
Recall from the first lesson:
1
lim− = −∞
𝑥→0 𝑥
1
lim+ = ∞
𝑥→0 𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
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lim = 𝐷𝑁𝐸
𝑥→0 𝑥
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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
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Example 2
x2  x  2
f x  
Let
x2 1
Identify all holes and asymptotes and find the left and right
hand limits as x approaches the vertical asymptotes.
(𝑥 + 2)(𝑥 − 1) 𝑥 + 2
𝑓 𝑥 =
=
(𝑥 + 1)(𝑥 − 1) 𝑥 + 1
𝐻𝑜𝑙𝑒: (1, 1.5)
𝑉𝐴: 𝑥 = −1
𝐻𝐴: 𝑦 = 1
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 2 (continued)
x2  x  2
f ( x) 
2
x 1
Vertical Asymptote
Hole
Horizontal
Asymptote
𝑥+2
lim −
= −∞
𝑥→−1 𝑥 + 1
𝑥+2
lim +
=∞
𝑥→−1 𝑥 + 1
Barnett/Ziegler/Byleen Business Calculus 12e
𝑥+2
lim
= 𝐷𝑁𝐸
𝑥→−1 𝑥 + 1
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Example 3
1
Let 𝑓 𝑥 = (𝑥 − 2)2
Identify all holes and asymptotes and find the left and right
hand limits as x approaches the vertical asymptotes.
1
𝑓 x =
𝑁𝑜 𝐻𝑜𝑙𝑒𝑠
(𝑥 − 2)2
𝑉𝐴: 𝑥 = 2
𝐻𝐴: 𝑦 = 0
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 3 (continued)
1
lim
=∞
𝑥→2− (𝑥 − 2)2
1
1
lim+
= ∞ lim
=∞
2
2
𝑥→2 (𝑥 − 2)
𝑥→2 (𝑥 − 2)
Barnett/Ziegler/Byleen Business Calculus 12e
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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
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End Behavior Review
Even degree
Negative
leading
coefficient
Odd degree
Positive
leading
coefficient
Odd degree
Negative
leading
coefficient
Barnett/Ziegler/Byleen Business Calculus 12e
Even degree
Positive
leading
coefficient
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Polynomial Functions
 Ex 4: Evaluate each limit.
lim 𝑥 2 = ∞
𝑥→−∞
lim 𝑥 2 = ∞
𝑥→+∞
lim
𝑥→−∞
lim −3𝑥 4 = −∞
𝑥→−∞
lim −3𝑥 4 = −∞
𝑥→+∞
6𝑥 3
= −∞
lim 6𝑥 3 = ∞
𝑥→+∞
Barnett/Ziegler/Byleen Business Calculus 12e
lim −5𝑥 3 = ∞
𝑥→−∞
lim −5𝑥 3 = −∞
𝑥→+∞
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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 𝑓 𝑥 = ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒 𝑣𝑎𝑙𝑢𝑒
𝑥→±∞
Barnett/Ziegler/Byleen Business Calculus 12e
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Rational Functions
 Ex 5: Evaluate
lim 𝑓 𝑥
𝑥→±∞
5
𝑓 𝑥 =
𝑥+2
𝐻𝐴: 𝑦 = 0
Because the degree of the
numerator < degree of the
denominator.
lim 𝑓 𝑥 = 0
𝑥→∞
lim 𝑓 𝑥 = 0
𝑥→−∞
Barnett/Ziegler/Byleen Business Calculus 12e
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Rational Functions
 Ex 6: Evaluate
3x 2  5 x  9
y
2x2  7
lim 𝑓 𝑥
𝑥→±∞
3
𝐻𝐴: 𝑦 =
2
Because the degree of the
numerator = degree of the
denominator.
3
lim 𝑓 𝑥 =
𝑥→∞
2
3
lim 𝑓 𝑥 =
𝑥→−∞
2
Barnett/Ziegler/Byleen Business Calculus 12e
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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
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Rational Functions
 Ex 7: Evaluate
lim 𝑓 𝑥
𝑥→±∞
2𝑥 5 − 𝑥 3 − 1
𝑓 𝑥 = 3
6𝑥 + 2𝑥 2 − 7
𝐻𝐴: 𝑁𝑜𝑛𝑒
Because the degree of the
numerator > degree of the
denominator.
2𝑥 5
𝑥2
lim
= lim
= ∞
𝑥→∞ 6𝑥 3
𝑥→∞ 3
2𝑥 5
𝑥2
lim
= lim
=∞
𝑥→−∞ 6𝑥 3
𝑥→−∞ 3
Barnett/Ziegler/Byleen Business Calculus 12e
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Rational Functions
 Ex 8: Evalaute
lim 𝑓 𝑥
𝑥→±∞
5𝑥 6 + 3𝑥
𝑓 𝑥 = 5
2𝑥 − 𝑥 − 5
𝐻𝐴: 𝑁𝑜𝑛𝑒
5𝑥
5𝑥 6
= ∞
lim 5 = lim
𝑥→∞ 2
𝑥→∞ 2𝑥
5𝑥
5𝑥 6
= −∞
lim
= lim
5
𝑥→−∞ 2
𝑥→−∞ 2𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
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Homework
Barnett/Ziegler/Byleen Business Calculus 12e
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Chapter 3
Limits and the
Derivative
Section 2
Infinite Limits and
Limits at Infinity
(Part 2)
Objectives for Section 3.2
Infinite Limits and Limits at Infinity
 The student will be able to solve applications involving limits.
Barnett/Ziegler/Byleen Business Calculus 12e
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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.
A. Assuming the total cost per day is linearly related to the
number of boards made per day, write an equation for the
cost function.
B. Write the equation for the average cost function.
C. Graph the average cost function: 𝑥: 1, 30 𝑦: [0, 500]
D. What does the average cost per board approach as
production increases?
Barnett/Ziegler/Byleen Business Calculus 12e
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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
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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
240𝑥 + 300
𝐶 𝑥 =
𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
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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]
Average
cost per
day
Number of surfboards
Barnett/Ziegler/Byleen Business Calculus 12e
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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?
240𝑥 + 300
𝐶 𝑥 =
𝑥
Average
cost per
day
Number of surfboards
As the number of boards increases, the
average cost approaches $240 per board.
Barnett/Ziegler/Byleen Business Calculus 12e
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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:
5𝑡 𝑡 + 50
𝐶 𝑡 = 3
𝑡 + 100
A. What is the drug concentration after 2 hours?
lim 𝐶(𝑡)
B. Evaluate and interpret the meaning of the limit: 𝑡→∞
Barnett/Ziegler/Byleen Business Calculus 12e
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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:
5𝑡 𝑡 + 50
𝐶 𝑡 = 3
𝑡 + 100
 What is the drug concentration after 2 hours?
5(2) 2 + 50
𝐶 2 =
≈ 4.8
3
2 + 100
After 2 hours, the concentration of
the drug is 4.8 mg/ml.
Barnett/Ziegler/Byleen Business Calculus 12e
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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:
5𝑡 𝑡 + 50
𝐶 𝑡 = 3
𝑡 + 100
lim 𝐶(𝑡)
 Evaluate and interpret the meaning of the limit: 𝑡→∞
5𝑡 𝑡 + 50
lim 3
=0
𝑡→∞ 𝑡 + 100
As time passes, the drug concentration
approaches 0 mg/ml.
Barnett/Ziegler/Byleen Business Calculus 12e
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Homework
Barnett/Ziegler/Byleen Business Calculus 12e
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