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# AMS/ECON 11A , Week 1 Class Notes AMS11A

Marketplace > University of California - Santa Cruz > Economics > AMS11A > AMS ECON 11A Week 1 Class Notes
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Covers section 10.1, focuses on limits
COURSE
Math Methods for Economics
PROF.
Jonathan Katznelson
TYPE
Class Notes
PAGES
3
WORDS
CONCEPTS
Limits, Math, Econ, Economics, AMS, Econ11A
KARMA
25 ?

## Popular in Economics

This 3 page Class Notes was uploaded by Adriana Chavez on Tuesday September 27, 2016. The Class Notes belongs to AMS11A at University of California - Santa Cruz taught by Jonathan Katznelson in Fall 2016. Since its upload, it has received 11 views. For similar materials see Math Methods for Economics in Economics at University of California - Santa Cruz.

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Date Created: 09/27/16
Tuesday, September 27, 2016 AMS/ECON 11A - Week One Class Notes Limits • Reading Due: Section 10.1 • Homework: 10.1 #3, 5, 8, 11, 15, 18, 21, 25, 28, 36, 37, 40, 43 - Class Notes Compounding Intrest - Intro to Limits • - ex. Intrest rate: r = 5% per year , investing \$1,000 for t years • Compounding Annually → value increases by 5% each year S(1) = 1000 + 1000(0.05) = \$1050 after one year • Compounding Monthly: S(1) = 1000 (1 + (0.05/12)) ← After one month S(1) = 1000 (1 + (0.05/12)) ← After two months S(1) = 1000 (1 + (0.05/12)) ← After three months […] S(1) = 1000 (1 + (0.05/12)) 12← After twelve months (one year) = \$1051.16 after one year Compounding Daily: • 365 S(1) = 1000 (1 + (0.05/365)) ≈ \$1051.267 after one year Compounded 1000 times/year • 1000 S(1) = 1000 (1 + (0.05/1000)) ≈ \$1051.269 after one year µ = length of compounding period k = number of periods / year k = 1 / µ • Question: As µ gets smaller (and k gets bigger), what happens to the amount of money after one year? 1 Tuesday, September 27, 2016 S 1/µ = 1000 (1 + 0.05µ) 1/µ← where µ is approaching 0 µ = 1: S(1) = 1050 µ = 12: S(1) = 1051.16 µ = 360: S(1) = 1051.267 µ = 1000: S(1) = 1051.269 • Observations: As µ approaches 0, S1/µ(1) approaches 1051.27109 S 1/0 = 1000 (1 + 0(0.05)) 1/0← not deﬁned • ex: f(x) = x - 4 x - 2 f(x) is not defined when x = 2 Graph of f(x) = x2 - 4 For all values approaching 2 x - 2 - f(1) = (-3 / -1) = 3 (but ≠ 2), the function works. 5 - f(3) = ( 5 / 1) = 5 - f(1.9) = 3.9 4 - f(1.99) = 3.99 3 - f(2.1) = 4.1 1 2 3 - f(2.01) = 4.01 It appears that as x → 2, f(x) → 4 ✴ Deﬁnition: • The limit, as x → a of f(x) is equal to L if f(x) becomes closer and closer to L as x gets closer to a. x ≠ a • If there is no number L, we say that the limit does not exist. ✴ Notation: lim (f(x)) = L • x → a - ex. #1: lim 1000(1 + 0.05µ) 1/µ= 1051.2709 Don’t just plug 9 in. x → 0 When x ≈ 9, √x ≈ 3 - ex. #2: lim x - 4 = 4 x → 2 x - 2 Proof: x - 9 = (√(x) - 3) (√(x) + 3) - ex. #3: x → 9x = 3 √(x) - 3 = x - 9 √(x) + 3 2 Tuesday, September 27, 2016 - Back to ex. #2: • f(x) = x - 4 x - 2 (x - 4) = (x-2)(x+2) 2 x - 4 = (x-2)(x+2) = x + 2if ≠ 2 x - 2 x - 2 2 If x ≈ 2, then x - 4 = x + 2 ≈ 2 + 2 = 4 x - 2 ✴ Rules for computing limits:(shortcuts) 1. lim C = C x → a 2. x → a = a 3. I Fx → a(f(x)) = LAND lim (g(x)) = M THEN x → a(f(x) ± g(x)) = L ± M x → a 4. I F lim (f(x)) = LAND lim (g(x)) = M THEN lim (f(x) ⦁ g(x)) = L⦁ M x → a x → a x → a 5. I Fx → a(f(x)) = LAND x → a(g(x)) = M THEN x → af(x) ÷ g(x)) = L ÷ Mif M ≠ 0 6. lx → a= a if a ≠ 0; true for any k for which ais defined • ex. lx → 9 = 3 = lx → 9= lim x → 91/2= 3 7. IF lim (f(x)) = LAND lim (g(x)) = M THEN lim (f(x)) = M x → a x → L x → a 8. IF g(x) = f(x) for all x ≠ THEN ,x → a(f(x)) =x → ag(x))OR , neither one exists 3

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