1. 8.7: Indeterminate Forms and L’Hôpital’s Rule
5, 9, odd (pick 5), odd (pick 6), 65,
Learning Objectives:
Recognize limits that produce indeterminate forms
Apply L’Hopital’s Rule to evaluate a limit

2. This will be the last PowerPoint I assemble for AP Calculus AB @ SKHS
Professional Note
This will be the last PowerPoint I assemble for AP Calculus SKHS

3. Concept 1A: Indeterminate Form
Info Only
0 0 and ∞ ∞ are both considered indeterminate forms because they do not guarantee a limit exists

4. Concept 1A: Indeterminate Form
Info Only
0 ∞ is not really an issue as it equals zero ∞ 0 is undefined but not indeterminate

5. Concept 1A: Indeterminate Form
The case:
lim 𝑥→−2 𝑥 2 −4 𝑥+2

6. Concept 1A: Indeterminate Form
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lim 𝑥→−2 𝑥 2 −4 𝑥+2
= lim 𝑥→−2 (𝑥+2)(𝑥−2) 𝑥+2
= lim 𝑥→−2 𝑥−2
=−2−2
=−4

7. Concept 1A: Indeterminate Form
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lim 𝑥→−2 𝑥 2 −4 𝑥+2

8. Concept 1A: Indeterminate Form
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The ∞ ∞ case:
lim 𝑥→∞ 2 𝑥 𝑥 2 −1
= ∞+1 ∞−1

9. Example 1A: Evaluating Indeterminate Forms
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Example 1A: Evaluating Indeterminate Forms
In chapter 1…
Lesson 1.5
…Concept 1
= lim 𝑥→∞ 𝑥 3 𝑥 4 − 4𝑥 𝑥 𝑥 4 𝑥 𝑥 4
= lim 𝑥→∞ 1 𝑥 − 4 𝑥 𝑥 4
lim 𝑥→∞ 𝑥 3 −4𝑥 2 𝑥 4 +5
lim 𝑥→∞ 𝑥 3 −4𝑥 2 𝑥 4 +5
= 0−0 2+0
= 0 2 =0

10. Example 1A: Evaluating Indeterminate Forms
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= lim 𝑥→∞ 4𝑥 𝑥 2 − 3 𝑥 𝑥 2 𝑥 𝑥 2
= lim 𝑥→∞ 4 𝑥 − 3 𝑥 𝑥 2
lim 𝑥→∞ 4𝑥−3 5 𝑥 2 +1
= 4 ∞ − 3 ∞ ∞
= 0−0 5+0 =0

11. Student Led Example 1A
None

12. Concept 1B: Le Frenchy’s Rule
L’Hopital’s Rule
lim 𝑥→𝑐 𝑓 𝑥 𝑔(𝑥) = lim 𝑥→𝑐 𝑓 ′ 𝑥 𝑔 ′ 𝑥

13. Concept 1B: Le Frenchy’s Rule
𝑓′ 𝑥 = 𝑥 2 −4𝑥−1 𝑥−2 2
𝑓 𝑥 = 𝑥 2 +1 𝑥−2

14. Concept 1B: Le Frenchy’s Rule

15. Example 1B: Evaluating Indeterminate Forms – Polynomial
lim 𝑥→∞ 4𝑥−3 5 𝑥 2 +1
= lim 𝑥→∞ 𝑥
= 4 ∞ =0

16. Student Led Example 1B: Evaluating Indeterminate Forms – Polynomial
lim 𝑥→−2 𝑥 2 −3𝑥−10 𝑥+2 −7

17. Example 1C: Evaluating Indeterminate Forms – Natural Log
= lim 𝑥→1 3⋅ 1 𝑥 2𝑥
lim 𝑥→1 ln 𝑥 3 𝑥 2 −1
= lim 𝑥→1 3 ln 𝑥 𝑥 2 −1
= lim 𝑥→1 3 𝑥 ⋅ 1 2𝑥
= lim 𝑥→ 𝑥 2 =
= 3 2

18. Student Led Example 1C: Evaluating Indeterminate Forms – Natural Log
lim 𝑥→∞ ln 𝑥 4 𝑥 3

19. Example 1D: Evaluating Indeterminate Forms – Exponential
= lim 𝑥→∞ 3 𝑥 2 2𝑥𝑒 𝑥 2
lim 𝑥→∞ 𝑥 3 𝑒 𝑥 2
= lim⁡ 𝑥→∞ 6𝑥 2⋅ 𝑒 𝑥 2 +2𝑥(2𝑥 𝑒 𝑥 2 )

20. Example 1D: Evaluating Indeterminate Forms – Exponential
Sometimes you
can simplify
too far
= lim⁡ 𝑥→∞ 6𝑥 2⋅ 𝑒 𝑥 2 +2𝑥(2𝑥 𝑒 𝑥 2 )
= lim⁡ 𝑥→∞ 6𝑥 2 𝑒 𝑥 𝑥 2 𝑒 𝑥 2
= lim⁡ 𝑥→∞ 6𝑥 𝑒 𝑥 2 (2+4 𝑥 2 )
Let’s take it from here

21. Example 1D: Evaluating Indeterminate Forms – Exponential
𝑥 2 𝑒 𝑥 2
= lim⁡ 𝑥→∞ 6𝑥 2 𝑒 𝑥 2 ′ +4 𝑥 2 𝑒 𝑥 2 ′
𝑓 𝑥 = 𝑥 2 𝑔 𝑥 = 𝑒 𝑥 2 𝑓 ′ 𝑥 =2𝑥 𝑔 ′ 𝑥 =2𝑥 𝑒 𝑥 2 C B
Let’s take it from here A D
= lim⁡ 𝑥→∞ 6 4𝑥 𝑒 𝑥 𝑥⋅ 𝑒 𝑥 2 + 𝑥 2 ⋅2𝑥 𝑒 𝑥 2 A C B D

22. Example 1D: Evaluating Indeterminate Forms – Exponential
= lim⁡ 𝑥→∞ 6 4𝑥 𝑒 𝑥 𝑥⋅ 𝑒 𝑥 2 + 𝑥 2 ⋅2𝑥 𝑒 𝑥 2
= lim⁡ 𝑥→∞ 6 4𝑥 𝑒 𝑥 2 +8𝑥 𝑒 𝑥 𝑥 3 𝑒 𝑥 2
= lim⁡ 𝑥→∞ 𝑥 𝑒 𝑥 𝑥 3 𝑒 𝑥 2
= 6 ∞ =0

23. Student Led Example 1D: Evaluating Indeterminate Forms - Exponential
lim 𝑥→∞ 𝑒 𝑥 2 𝑥 ∞

24. CHALLENGE!
Explain why L’Hopital’s Rule does NOT work for this problem:
lim 𝑥→0 𝑒 𝑥 2 𝑥

25. Example 1D: Evaluating Indeterminate Forms – Trigonometric
Arctan?
Example 1D: Evaluating Indeterminate Forms – Trigonometric
lim 𝑥→1 arctan 𝑥 − 𝜋 4 𝑥−1
What the is Arctan?

26. Concept 1D: Inverse Trigonometric Functions
Info Only
arcsin 𝑥 = sin −1 𝑥 arccos 𝑥 = cos −1 𝑥 …and so on arcsin sin 𝑥 =𝑥 ..and so on

27. Concept 1D: Inverse Trigonometric Functions
Option 1: Go home, open your book to p.369 & 371 and copy down Theorem 5.16 and “Basic Dedifferentiation Rules for Elementary Functions” into your notes. These lessons are posted far enough in advance for you to behave proactively about your education

28. Concept 1D: Inverse Trigonometric Functions
Option 2: Google, find and print a sheet (ideally a PDF) which contains this information. Have Mr. Hansen look at it, initial it and you can use it on a test

29. Concept 1D: Inverse Trigonometric Functions
Option 3: Buy a Spark Chart or equivalent

30. Concept 1D: Inverse Trigonometric Functions
Not an Option: Mr. Hansen muddles around the classroom waiting for students to write things into their notebooks and watch as y’all get board waiting on the slow writers.

31. Example 1D: Evaluating Indeterminate Forms – Trigonometric
lim 𝑥→1 arctan 𝑥 − 𝜋 4 𝑥−1
tan −1 𝑢 ′ = 1 1+ 𝑢 2
𝑢=𝑥
lim 𝑥→ 𝑥 2 1
= lim 𝑥→ 𝑥 2
tan −1 𝑥 ′ = 1 1+ 𝑥 2
= 1 2

32. Concept 2: Other Forms
∞−∞
0⋅∞
1 ∞
0 0

33. lim 𝑥→∞ 𝑒 −𝑥 𝑥 ⇒ 𝑒 −∞ ∞ =0⋅∞ = lim 𝑥→∞ 𝑥 𝑒 𝑥 ⇒ ∞ 0 Example 2A: 0⋅∞
Info Only
lim 𝑥→∞ 𝑒 −𝑥 𝑥
⇒ 𝑒 −∞ ∞
=0⋅∞
= lim 𝑥→∞ 𝑥 𝑒 𝑥
⇒ ∞ 0
Turn products into quotients!

34. lim 𝑥→∞ 𝑥 𝑒 𝑥 = lim 𝑥→∞ 1 2 𝑥 ⋅𝑒 𝑥 = 1 2 ∞ ⋅𝑒 ∞ = 1 ∞⋅∞ =0
Example 2A: 0⋅∞
Info Only
lim 𝑥→∞ 𝑥 𝑒 𝑥
= lim 𝑥→∞ 𝑥 ⋅𝑒 𝑥
= 1 2 ∞ ⋅𝑒 ∞
= 1 ∞⋅∞ =0

35. lim 𝑥→∞ 1+ 1 𝑥 𝑥 ln lim 𝑥→∞ 1+ 1 𝑥 𝑥 = ln 𝑦 lim 𝑥→∞ ln 1+ 1 𝑥 𝑥 = ln 𝑦
Example 2B: 1 ∞
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lim 𝑥→∞ 𝑥 𝑥
Some rules we can apply to expressions come from equations
ln lim 𝑥→∞ 𝑥 𝑥
= ln 𝑦
lim 𝑥→∞ ln 𝑥 𝑥
= ln 𝑦

36. lim 𝑥→∞ ln 1+ 1 𝑥 𝑥 𝑥⋅𝑢= 𝑢 𝑥 −1 = ln 𝑦 Product lim 𝑥→∞ 𝑥 ln 1+ 1 𝑥
Example 2B: 1 ∞
Info Only
lim 𝑥→∞ ln 𝑥 𝑥
𝑥⋅𝑢= 𝑢 𝑥 −1
= ln 𝑦
Product
lim 𝑥→∞ 𝑥 ln 𝑥
= ln 𝑦
lim 𝑥→∞ ln 𝑥 𝑥 −1

37. Let 𝑢=1+ 1 𝑥 lim 𝑥→∞ ln 1+ 1 𝑥 𝑥 −1 𝑢 ′ =0− 𝑥 −2 ln 𝑢 ′ = 1 𝑢
Example 2B: 1 ∞
This guy first
lim 𝑥→∞ ln 𝑥 𝑥 −1
𝑢 ′ =0− 𝑥 −2
ln 𝑢 ′ = 1 𝑢
lim 𝑥→∞ 1 𝑢 ⋅𝑢′ 𝑥 −1 ′
1 𝑢 = 𝑥

38. = lim 𝑥→∞ 1 1+ 1 𝑥 ⋅− 𝑥 −2 − 𝑥 −2 lim 𝑥→∞ 1 𝑢 ⋅𝑢′ 𝑥 −1 ′
Example 2B: 1 ∞
= lim 𝑥→∞ 𝑥 ⋅− 𝑥 −2 − 𝑥 −2
lim 𝑥→∞ 1 𝑢 ⋅𝑢′ 𝑥 −1 ′
= lim 𝑥→∞ 𝑥

39. Example 2B: 1 ∞
lim 𝑥→∞ 𝑥
= ∞
= 1 1+0 =1

40. lim 𝑥→ 1 + 1 ln 𝑥 ⋅ 𝑥−1 𝑥−1 − 1 𝑥−1 ⋅ ln 𝑥 ln 𝑥
Example 2C: ∞−∞
lim 𝑥→ ln 𝑥 − 1 𝑥−1
lim 𝑥→ ln 𝑥 ⋅ 𝑥−1 𝑥−1 − 1 𝑥−1 ⋅ ln 𝑥 ln 𝑥
lim 𝑥→ 𝑥−1− ln 𝑥 ( ln 𝑥)(𝑥−1)

41. lim 𝑥→ 1 + 𝑥−1− ln 𝑥 ( ln 𝑥)(𝑥−1)
Example 2C: ∞−∞
𝑥−1− ln 𝑥 ′ =1−0− 1 𝑥 =1− 1 𝑥
lim 𝑥→ 𝑥−1− ln 𝑥 ( ln 𝑥)(𝑥−1)
ln 𝑥 ′ 𝑥−1 + ln 𝑥 𝑥−1 ′
1 𝑥 𝑥−1 + ln 𝑥 ⋅1

42. =UND = 0 1 1 0 +0 Example 2C: ∞−∞ lim 𝑥→ 1 + 1− 1 𝑥 1 𝑥 𝑥−1 + ln 𝑥=
=UND
Sometimes we need to differentiate twice

43. Example 2C: ∞−∞
lim 𝑥→ − 1 𝑥 1 𝑥 𝑥−1 + ln 𝑥
= lim 𝑥→ − 1 𝑥 1− 1 𝑥 + ln 𝑥
= lim 𝑥→ 𝑥 −2 𝑥 −2 + 𝑥 −1
= lim 𝑥→ − 𝑥 −1 1− 𝑥 −1 + ln 𝑥
= 1 −2 1 −2 + 1 −1 =
= 1 2

44. Student Led Example 2
None