Lecture 5: L’Hôpital’s Rule, Sequences, and Series

Part I: L’Hôpital’s Rule

Indeterminate forms How do we compute lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) when 𝑓 𝑎 =𝑔 𝑎 =0? Similarly, how do we compute lim 𝑥→∞ 𝑓(𝑥) 𝑔(𝑥) when lim 𝑥→∞ 𝑓 𝑥 = ±∞ and lim 𝑥→∞ 𝑔 𝑥 = ±∞ ?

Computing Indeterminate forms If 𝑓 𝑎 =𝑔 𝑎 =0, 𝑓 ′ 𝑎 and 𝑔′(𝑎) exist, and 𝑔′(𝑎)≠0 then lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = 𝑓 ′ (𝑎) 𝑔 ′ (𝑎) To see this, note that lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = lim ∆𝑥→0 𝑓 𝑎+∆𝑥 −𝑓(𝑎) 𝑔 𝑎+∆𝑥 −𝑔(𝑎) = lim ∆𝑥→0 𝑓 𝑎+∆𝑥 −𝑓(𝑎) ∆𝑥 ∙ lim ∆𝑥→0 ∆𝑥 𝑔 𝑎+∆𝑥 −𝑔(𝑎) = 𝑓 ′ (𝑎) 𝑔 ′ (𝑎) What happens if 𝑓 ′ 𝑎 = 𝑔 ′ 𝑎 =0?

L’Hǒpital’s Rule L’Hǒpital’s Rule: If 𝑓 𝑎 =𝑔 𝑎 =0 then lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = lim 𝑥→𝑎 𝑓 ′ (𝑥) 𝑔 ′ (𝑥) provided that the limit on the right exists and 𝑔′(𝑥)≠0 when 𝑥 is close to 𝑎 but not equal to it. L’Hǒpital’s Rule for limits at infinity: If lim 𝑥→∞ 𝑓 𝑥 = ±∞ and lim 𝑥→∞ 𝑔 𝑥 = ±∞ then lim 𝑥→∞ 𝑓(𝑥) 𝑔(𝑥) = lim 𝑥→∞ 𝑓 ′ (𝑥) 𝑔 ′ (𝑥) provided that the limit on the right exists and 𝑔′(𝑥)≠0 when 𝑥 is sufficiently large. A similar statement holds for limits at −∞.

L’Hǒpital’s Rule Examples lim 𝑥→∞ 2𝑥 2 +3𝑥+4 𝑥 2 = lim 𝑥→∞ 4𝑥+3 2𝑥 = lim 𝑥→∞ 4 2 =2 lim 𝑥→0 sin 𝑥 −𝑥 𝑥 3 = lim 𝑥→0 cos 𝑥 −1 3𝑥 2 = lim 𝑥→0 −sin 𝑥 6𝑥 = lim 𝑥→0 −cos 𝑥 6 = −1 6

Part II: Sequences and Series

Objectives Know how to find the limit of sequences Know how to compute geometric series and telescoping series Corresponding sections in Simmons: 13.2, 13.3

Sequences If we have an element 𝑥 𝑛 for every n, this is called a sequence Example: 1,2,3,4,… 𝑥 𝑛 =𝑛 Example: 1, 1 2 , 1 3 , 1 4 ,… 𝑥 𝑛 = 1 𝑛

Convergence of Sequences We say a sequence 𝑥 1 , 𝑥 2 ,… converges to L if lim 𝑛→∞ 𝑥 𝑛 =𝐿 Limits of sequences can be found in the same way as limits of functions. L’Hopital’s rule often works, sometimes other tools are needed. Example: If 𝑥 𝑛 = 𝑛 2 +2𝑛 2𝑛 2 +5 , lim 𝑛→∞ 𝑥 𝑛 = 1 2 Example: If 𝑥 𝑛 = 𝑛 2 +2𝑛 −𝑛, complete the square. 𝑥 𝑛 = 𝑛 2 +2𝑛 −𝑛= 𝑛+1 2 −1 −𝑛 𝑥 𝑛 ≈𝑛+1− 1 2 𝑛+1 −𝑛 so lim 𝑛→∞ 𝑥 𝑛 =1

Series A series is an expression of the form 𝑛=1 ∞ 𝑥 𝑛 = 𝑥 1 + 𝑥 2 + 𝑥 3 + 𝑥 4 +… Can be viewed as a sequence 𝑦 𝑛 = 𝑗=1 𝑛 𝑥 𝑗 Key example: Geometric series Geometric series have the form 1+𝑎+ 𝑎 2 + 𝑎 3 +… (times a constant)

Evaluating Series Geometric series can be evaluated as follows 1+𝑎+…+ 𝑎 𝑛−1 = 1−𝑎 𝑛 1−𝑎 𝑛=1 ∞ 𝑎 𝑛−1 = 1 1−𝑎 if and only if 𝑎 <1 Partial fractions and cancelling terms can also be useful 𝑛=1 ∞ 1 𝑛 2 +𝑛 = 𝑛=1 ∞ 1 𝑛 − 1 𝑛+1 = 𝑛=1 ∞ 1 𝑛 − 𝑛=2 ∞ 1 𝑛 =1 When terms cancel like this, it is called a telescoping series.