Question

So, I have my basic analysis final today, and quite honestly, I need help. So for my first question, can someone help me make an outline for an epsilon delta proof?

Answer 1

ok so this is what I have:
\(set~\epsilon>0\) \(Define~ \delta :=n\epsilon~~~~ (have~to ~include~max~here~somehow?)~~s.t.~~~|x-c|<\delta \)
Then \[|f(x)-L| \le\]...\[~~~~~~~|x-c|\le \epsilon \]

Answer 2

@zzr0ck3r , Does that work? And the final should be epsilon/delta

Answer 3

tsk

Answer 4

if only dan815 were here to save us

Answer 5

the way to make the perfect user name is take your hand spread it open wide and scramble it across the keyboard

Answer 6

the ... is algebra to get to the point where we have delta

Answer 7

\(\forall x\in D \ \forall \epsilon >0 \ \exists\delta >0 \ \forall x\in D \ (|x-a|<\delta \implies|f(x)-f(a)|<\epsilon )\)

Answer 8

This is continuity

Answer 9

but is that ok for just limit?

Answer 10

ahh well we need a few quantifiers

Answer 11

ok, so should I use set or for every for epsilon, or does it matter?

Answer 12

\( \forall \epsilon >0 \ \exists\delta >0 \ \forall x\in D-\{c\} \ (|x-c|<\delta \implies|f(x)-L|<\epsilon )

\)

Answer 13

what is the x an element of D for?

Answer 14

ohoh oh nvm

Answer 15

the -{c} might not be in the definition you use

Answer 16

It does not matter

Answer 17

ok so where does the delta being in the max{} belong?

Answer 18

well normally we use min not max

Answer 19

since often we need to have multiple conditions on |x-c|

Answer 20

oh oops...

Answer 21

you have it backwards, normally its like this

\(|x-c|\implies ....\) Since delta is less than blah \(.......\) thus \(|f(x)-L|< \epsilon\)

Answer 22

I guess I never understood how to write the bounds. Like if it is from {1, epsilon/2},

Answer 23

Say we want to prove that the limit of f(x) = 3x as x goes to 3

Answer 24

ok so first we guess L and do the side work to find delta right?

Answer 25

right, L = 9

Answer 26

so then \[|3x-9| \le 3|x-3|\]\[~~~~~~~~~~~~\le \epsilon/3\]

Answer 27

so we define delta as epsilon over 3

Answer 28

then we write up the whole thing.... right?

Answer 29

we want|

\(|3x-9|<\epsilon\)

so

\(|3x-3^2| = 3|x-3|\)


pf:

Let \(\epsilon > 0\)
Choose delta such that \(\delta <\frac{\epsilon}{3}\)

Then \(|x-3|<\delta \implies |x-3|<\frac{\epsilon}{3}\implies |f(x)-9|=|3x-9|=3|x-3|<\epsilon\)

Answer 30

Sorry I didn't read what you wrote

Answer 31

yes

Answer 32

Ok but what about something like f(x)=x^2

Answer 33

then we have to use the conjugate trick

Answer 34

then we will have more than one delta

Answer 35

nah that's with sqrt

Answer 36

ok so \[|x^2-c^2|\le |x+c||x-c|\]
Then we have to bound x+c

Answer 37

but on the reals, we can't make a bound... can we?

Answer 38

\(|x^2-3^2|=|x-3||x+3|\)

So we will do a few things

Suppose \(|x-3|<1\) then \(x-3<1\implies x+3<7\)

Answer 39

yes

Answer 40

ok so wlog we can assume that |x+c|<1

Answer 41

now we have choose \(\delta , \min(\frac{\epsilon}{7}, 1)\)

Answer 42

\(\delta < \min(\frac{\epsilon}{7}, 1)

\)

Answer 43

how did you choose the 7?

Answer 44

I wanted |x+3| < something

Answer 45

the choice of 1, led to the 7

Answer 46

what if you had a general case?

Answer 47

?

Answer 48

like instead of 3, you had c?

Answer 49

and we want to get get \(|x+c|<\)something?

Answer 50

right

Answer 51

\(|x-c|<1\implies x-c<1\implies x+c<1+2c\implies |x+c|<1+2c

\)

Answer 52

so then we would use 1+2c?

Answer 53

well epsilon over 1+2c?

Answer 54

well, then you got to remember, that you will be replacing |x+3| with that in the string of inequalities, so you will have, most likely, something like \(\delta < \min(1, \frac{\epsilon}{1+2c})\)

Answer 55

no, delta < 1 gives us |x+3|< 1+2c

Answer 56

errrrr

Answer 57

replace the 3's with c's lol

Answer 58

ok, so for the general, we get 1+2C because we bounded |x-c|, yes?

Answer 59

yes

Answer 60

ok, and the min is placed directly after we define delta in the prffof?

Answer 61

oh goodness, proof**

Answer 62

lol

Answer 63

we define delta with min

Answer 64

ohhhh ok

Answer 65

in most cases....

Answer 66

you could equaly choose deta = min{}

Answer 67

not <....

Answer 68

... ya know, conceptually I get this stuff, but for some reason, this one proof, has never sunk in

Answer 69

we have to define delta>0 first though, right?

Answer 70

yeah we should say \(0< \delta < \min \{....\}\)

Answer 71

we can say choose a delta in that interval, or use = instead of < on the rhs

Answer 72

the stuff we pick inside the min must always be positive...

Answer 73

What else you got? its 445 am ;)

Answer 74

no one should be doing epsilon delta proofs in this hour

Answer 75

ok and lol it's 745 here :)

Answer 76

And, that settles ep delta question for limits, the only difference for continuity os we use f(c) and then I'll open a new question for my second query

Answer 77

thanks :) if you have to go, I understand :)

Answer 78

lol

Answer 79

god I love the west coast

Answer 80

haha east COAAASSSSSSTTT is where it's at bro

Answer 81

but unfortunately crappy teachers also call this place home o.o

Answer 82

lol 2pac 4 life son

Answer 83

jk ;P

Answer 84

lol