Cornell Notes Section 1.2 Day 1 Section 1.2 Day 2 Section 1.3 Day 1 Write a summary of the lesson Section 1.2 Day 2 Write 2 questions (L1-L3) Section 1.3 Day 1 Read through highlight, underline, add “?”

Exit Ticket for Feedback Let 𝐥𝐢𝐦 𝒙→𝟏 𝒇(𝒙) =−𝟏 and 𝐥𝐢𝐦 𝒙→𝟏 𝒈(𝒙) =𝟑. Find the following: 1) lim 𝑥→1 4𝑓 𝑥 +9 = 3) lim 𝑥→1 𝑔 𝑥 2 = 2) lim 𝑥→1 𝑔(𝑓 𝑥 ) = 4) lim 𝑥→1 𝑔 𝑥 𝑓 𝑥 =

Section 1.3 Day 2 Evaluating limits analytically AP Calculus AB

By the end of this lesson, you should… Be familiar with basic limit properties Determine the limit of a composition of functions Determine the limit of a function using direct substitution Determine the limit of a function by simplification Determine the limit of a function by rationalizing Determine the limit of a function using special trigonometric relationships Determine the limit of a function using Squeeze Theorem Define and recognize indeterminate forms

Method 1: Direct Substitution lim 𝑥→𝑐 𝑓(𝑥) =𝑓(𝑐) In most cases, 𝑓(𝑥) is a continuous function, and the limit value is precisely the function value. However, it still holds that the function value does not affect the limit value. Direct substitution is usually the first method used to find a limit analytically 𝑐

Method 1: Direct Substitution Example 1: lim 𝑥→2 4 𝑥 2 +3 = 19 Example 2: lim 𝑥→1 𝑥 2 +𝑥+2 𝑥+1 = 2 Example 3: lim 𝑥→0 2+ 𝑒 𝑥 2 = 3 Example 4: lim 𝑥→3 3 2 𝑥 2 −10 = 2 Example 5: lim 𝑥→ 𝜋 𝑥 cos 𝑥 = −𝜋

Method 2: Simplification In your groups, discuss why the following two limits are the same. Support your answer with mathematics lim 𝑥→−3 𝑥 2 +𝑥−6 𝑥+3 = lim 𝑥→−3 𝑥−2 =−5

Method 2: Simplification lim 𝑥→−3 𝑥 2 +𝑥−6 𝑥+3 = lim 𝑥→−3 𝑥−2 =−5 Let’s take a look at this graphically

Method 2: Simplification Notice that if we use the method of direct substitution, lim 𝑥→−3 𝑥 2 +𝑥−6 𝑥+3 results in the form 0 0 . This is an indeterminate form.

Indeterminate Forms An indeterminate form does not have enough information to conclude the value of a limit Some indeterminate forms can look like the following: 0 0 , ∞ ∞ , 0 0 , 0 ∞ , ∞ 0

Indeterminate Forms Let’s take a closer look at 0 0 . What does this really mean? 𝑥= 0 0 : 𝑥∙0=0 Notice that any possible number could satisfy the “x” in the above equation. Because of this, we cannot determine what “x” is, and we denote it as an indeterminate form.

Indeterminate Forms When taking the limit by direct substitution, if an indeterminate form occurs, another method needs to be used. Sometimes, the function needs to be rewritten into another equivalent function before analyzing the limit. One way to do this is by simplification. We will look more closely at this in the next slide

Method 2: Simplification If you result in an indeterminate form from direct substitution, look for ways to simplify the function first then use direct substitution. This is the 2nd method to finding a limit, and it is important to be proficient in factoring. Note: Simplification may change the function slightly. In most cases, it may change certain function values. However, it should not change the basic shape of the graph of the function, which is all we really need.

Method 2: Simplification Example 1: lim 𝑥→2 𝑥 2 −4 𝑥−2 = 4 Example 2: lim 𝑥→1 𝑥−1 𝑥 2 −1 = 1 2 Example 3: lim 𝑥→𝑎 𝑥 4 − 𝑎 4 𝑥−𝑎 = 4 𝑎 3

Method 3: Rationalizing This method is a very specific type of simplification that does not involve standard factoring skills. Primarily used when there is a complex square root somewhere in the function It’s process is to multiply by a conjugate function to both the numerator and the denominator

Method 3: Rationalizing Example 1: lim 𝑥→0 𝑥+1 −1 𝑥 Notice that direct substitution gives us an indeterminate form and there is no way to simplify. lim 𝑥→0 𝑥+1 −1 𝑥 lim 𝑥→0 𝑥+1 −1 𝑥 ∙ ( 𝑥+1 +1) ( 𝑥+1 +1) = lim 𝑥→0 𝑥+1 2 − 1 2 𝑥∙ 𝑥+1 +1 = lim 𝑥→0 𝑥+1−1 𝑥∙ 𝑥+1 +1 = lim 𝑥→0 𝑥 𝑥∙ 𝑥+1 +1 = lim 𝑥→0 1 𝑥+1 +1 = 1 0+1 +1 = 1 2

Method 3: Rationalizing Example 2: lim 𝑥→4 𝑥+5 −3 𝑥−4 = 1 6 lim 𝑥→4 𝑥+5 −3 𝑥−4 = lim 𝑥→4 𝑥+5 −3 𝑥−4 ∙ 𝑥+5 +3 𝑥+5 +3 = lim 𝑥→4 (𝑥+5)−9 𝑥−4 𝑥+5 +3 = lim 𝑥→4 𝑥−4 𝑥−4 𝑥+5 +3 = lim 𝑥→4 1 𝑥+5 +3 = 1 9 +3 = 1 6

Exit Ticket for FEedback 1. lim 𝑥→3 𝑥+1 −2 𝑥−3 = 2. lim 𝑥→0 10 𝑥 −1 𝑥+2 = 3. lim 𝑡→2 𝑡 2 −3𝑡+2 𝑡 2 −4 =