8.7: Indeterminate Forms and L’Hôpital’s Rule 5, 9, 11-41 odd (pick 5), 43-59 odd (pick 6), 65, 91. 93 Learning Objectives: Recognize limits that produce indeterminate forms Apply L’Hopital’s Rule to evaluate a limit

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 AB @ SKHS

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

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

Concept 1A: Indeterminate Form The 0 0 case: lim 𝑥→−2 𝑥 2 −4 𝑥+2

Concept 1A: Indeterminate Form Info Only 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

Concept 1A: Indeterminate Form Info Only lim 𝑥→−2 𝑥 2 −4 𝑥+2

Concept 1A: Indeterminate Form Info Only The ∞ ∞ case: lim 𝑥→∞ 2 𝑥 2 +1 3 𝑥 2 −1 = ∞+1 ∞−1

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

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

Student Led Example 1A None

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

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

Concept 1B: Le Frenchy’s Rule

Example 1B: Evaluating Indeterminate Forms – Polynomial lim 𝑥→∞ 4𝑥−3 5 𝑥 2 +1 = lim 𝑥→∞ 4 10𝑥 = 4 ∞ =0

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

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→1 3 2 𝑥 2 = 3 2 1 2 = 3 2

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

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 )

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 2 +4 𝑥 2 𝑒 𝑥 2 = lim⁡ 𝑥→∞ 6𝑥 𝑒 𝑥 2 (2+4 𝑥 2 ) Let’s take it from here

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 2 +4 2𝑥⋅ 𝑒 𝑥 2 + 𝑥 2 ⋅2𝑥 𝑒 𝑥 2 A C B D

Example 1D: Evaluating Indeterminate Forms – Exponential = lim⁡ 𝑥→∞ 6 4𝑥 𝑒 𝑥 2 +4 2𝑥⋅ 𝑒 𝑥 2 + 𝑥 2 ⋅2𝑥 𝑒 𝑥 2 = lim⁡ 𝑥→∞ 6 4𝑥 𝑒 𝑥 2 +8𝑥 𝑒 𝑥 2 +8 𝑥 3 𝑒 𝑥 2 = lim⁡ 𝑥→∞ 6 12𝑥 𝑒 𝑥 2 +8 𝑥 3 𝑒 𝑥 2 = 6 ∞ =0

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

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

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?

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

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

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

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

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.

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→1 1 1+ 𝑥 2 1 = lim 𝑥→1 1 1+ 𝑥 2 tan −1 𝑥 ′ = 1 1+ 𝑥 2 = 1 2

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

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!

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→∞ 1 2 𝑥 ⋅𝑒 𝑥 = 1 2 ∞ ⋅𝑒 ∞ = 1 ∞⋅∞ =0

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 ∞ Info Only lim 𝑥→∞ 1+ 1 𝑥 𝑥 Some rules we can apply to expressions come from equations ln lim 𝑥→∞ 1+ 1 𝑥 𝑥 = ln 𝑦 lim 𝑥→∞ ln 1+ 1 𝑥 𝑥 = ln 𝑦

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 1+ 1 𝑥 𝑥 𝑥⋅𝑢= 𝑢 𝑥 −1 = ln 𝑦 Product lim 𝑥→∞ 𝑥 ln 1+ 1 𝑥 = ln 𝑦 lim 𝑥→∞ ln 1+ 1 𝑥 𝑥 −1

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 1+ 1 𝑥 𝑥 −1 𝑢 ′ =0− 𝑥 −2 ln 𝑢 ′ = 1 𝑢 lim 𝑥→∞ 1 𝑢 ⋅𝑢′ 𝑥 −1 ′ 1 𝑢 = 1 1+ 1 𝑥

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

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

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

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→ 1 + 𝑥−1− ln 𝑥 ( ln 𝑥)(𝑥−1) ln 𝑥 ′ 𝑥−1 + ln 𝑥 𝑥−1 ′ 1 𝑥 𝑥−1 + ln 𝑥 ⋅1

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

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

Student Led Example 2 None