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## Calculus week 2 Lecture and book 2.1- 2.4 sections notes

by: Brenna Graham

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6

# Calculus week 2 Lecture and book 2.1- 2.4 sections notes 191g

Marketplace > New Mexico State University > Math > 191g > Calculus week 2 Lecture and book 2 1 2 4 sections notes
Brenna Graham
NMSU
GPA 3.9

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Extensive notes for 8/23/16 and 8/25/16 lectures, links to more info, and supplemented by book notes. Is 2.1- 2.4 sections.
COURSE
Calculus I
PROF.
Dr. Andres Contreras
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Calc, Calculus, Limits, continuity, Math, Mathematics, college, Squeeze Theorem
KARMA
25 ?

## Popular in Math

This 6 page Class Notes was uploaded by Brenna Graham on Sunday August 28, 2016. The Class Notes belongs to 191g at New Mexico State University taught by Dr. Andres Contreras in Fall 2016. Since its upload, it has received 14 views. For similar materials see Calculus I in Math at New Mexico State University.

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Date Created: 08/28/16
Lecture 2  Left off last time :  limf(x) = L  If for every ε> 0 we can find δ > 0  x→c 0<| x­c|< f(x)  |f(x)−L|  < ε    EXAMPLE 1:  lim(3x + 1)    x→4 f(x) = 3x + 1    LET ε > 0    f(x) − |3| < ε  where 0< |x­c| <δ   |3x+1­13|=  |3x­12|  3|x­4|  0< 3|x­4 |<ε    ε |x­4|< 3     δ = 3    One Sided Limits:  From the left to right   x→c−f(x)    ε < 0    |f(x)− L < ε    δ > 0    x ε(c− δ,c)    From right to left  lim f(x)    x→c+ x ε(c,c+ δ)    |x− c| < δ if and only if  x ε(c −δ,cδ)    If x→c− = lx→c+ L    Then  limf(x) = L    x→c EXAMPLE   x f(x) = |x| R\{0}   limf(x) =?    x→0 x→0− f(x) =?    Doesn’t exist from both sides  at the same time.  Infinite limits  x→cf(x) =+ ∞  f(if  increases without bounds, as  x → c    Vertical Asymptote at x=c if at both x→ c+ and x→ c­ it approaches ­/+∞     Example:  1 f(x) = x−5  definition  xεR/{5}            Properties of Limits:  1. If x→cf(x)  +  x→cG(x) exist then lx→cf(x)+ g(x)) exists  and same for the  reverse  2. x→ck f(x) = klx→c(x)  assuming k is a constant  3. limf(x)× g(x) = limf(x)  × limg(x)    x→c x→c x→c lim f(x) 4. lim f(x)= x→c if x / 0    x→c g(x) x→c g(x) 5. h is positive  a. lim[f(x)]  = (limf(x))   h x→c x→c b. lim f(x) = n limf(x) assuming lim   is positive  x→c√ √ x→c  →  6. Identities  a. limk = k   k→c b. limx = c   x→c p p c. limx  = c  assume c>0 if q is even  x→c Example:  limx  − 2x+ 1 =?  x→2 limx  + lim(− 2)x  + 1  x→2 x→2 (limx)  + (−2)limx  + 1  x→2 x→2 32 − 4 +1 = 29    If the equation is continuous just plug c in   It is continuous if (a,b) and let c (a,b)  AKA:  limf(x) = f(c)   x→c TA NOTES:  2.1  1. Average rate of change  f(x) on (a,b)  Δf(x)= f(b)−f(a  Δx b−a 2. Instantaneous rate of change  Δf(x) lim Δx   Δx→o Example:  Δx Δy on [2,6), y=4x­9  Solution  Δx = (4(5)+9)−(8+9)= 29−17 = 12 =​   Δy 5−2 3 3 Find instant rate of change at x=2 f(x) = 4x +9      [2, 2.01]  [2, 2.005)  [2, 2.001)  Δf(x) f(b)−f(a) Δf(x) f(b)−f(a) Using: Δx = b−a   Using: Δx = b−a   Using: f(2.01)−f(2) f(2.005)−f(2) Δf(x)= f(b)−f(a  2.01−2 =  2.005−2 =  f(2.001)−f(2) (4(2.01)+9)−(4(2)+9) (4(2.005)+9)−(4(2)+9) 2.001−2 =  0.01   0.005   (4(2.001)+9)−(4(2)+9) =0.04 = 4  = 0.02 = 4   0.001   0.01 0.005 0.004 = 0.001 = 4    Rate of change is (via squeeze theorem) = 4  Lecture 3  2.4 limits + continuity  If limf(x) = f(c)   thus limit exists if equal at both sides  x→c If f(x)is continuous at I then continuous on line L  lim f(x) = f(a)  f(x)   I =[a,b) If on (a,b) and  x→a+/− if  is continuous at point a.    Example  2 f(x) = x     2 limx    x→1 Continuous functions:  1/n 1. y = x    2.  y = sinx, y = cosx,   3. y = b   if b > 0, b = / 1    4. y = log  x if x > 0, b > 0, b = / 1    b If function  and gare continuous at x=c then following cont.   I. f + g or − g    II. kf, k is a constant   III. fg   IV.  f/g if c) = / 0    V. (even if  isn’t continuous)  F(x) = f(g(x))  is continuous  −1 If   is continuous with range R  f     is  with domain R  He showed proof for continuity but I’m lazy and it is not necessary for you to  learn. It’s in linear algebra at khan academy. All of the above’s proofs are at:  http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx  Evaluate limits by  1. Definitions  2. By determining continuity and plugging in  3. Reduce through known limits or squeeze theorem. Shown in Lecture 2  TA section.    h Should be known by student for use: lim e  −1 = 1   h→0 h

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