Question

Epsilon Delta Definition of the Limit. Would anybody mind taking the effort to help me understand this proof? I watched a couple of You tube videos but they didn't really help to relieve me from the confusion I have.

I would greatly appreciate it,
QRA

Answer 1

an example would be best

Answer 2

the e,d definition tends to work backwards. It assumes an area close to and surrounding the Limit. it then asks what area from the "x axis" will always fall within the area defined by epsilon

Answer 3

the area for your x value are defined as delta then

Answer 4

Instead of area you should probably call it a neighborhood or interval

Answer 5

Let's say we have prove Lim x -> 1 is 3 for the function f(x) = 3x(x-1)/(x-1)

Answer 6

i had considered calling it a swimming pool, but figured that might be over the top ;)

Answer 7

Could anyone mind showing me a picture to illustrate this?

Answer 8

the picture is equivalent to y=3x with a hole at 1

Answer 9

i am still wondering what "this proof" is in your question

Answer 10

What do you mean?

Answer 11

Ok if the example I provided wouldn't work, then would someone mind making an example and show me step-by-step as to how to work with this proof? If that is too much to ask then I would just like an explanation, and I'll see if I understand it.

Answer 12

Ok what confuses me is how did you get this:
" |x - 1| < d -> |f(x) - 3| < e "

Answer 13

thats just the definition.

Answer 14

What does it mean though?

Answer 15

In simpler terms

Answer 16

if you are trying to prove that the lim f(x) as x approaches c = L, then you use the definition:
|x-c|<d -->|f(x)-L|<e
for any epsilon, there must be a delta such that this statement is true.

Answer 17

when x is really close to 1 then f(x) will be really close to 3

Answer 18

if i was at my scanner id write some stuff out, but im at school unfortunately =/ im sure someone else can explain it.

Answer 19

this line
\[ |x - 1| < \delta -> |f(x) - 3| < \epsilon \] says in inglish that if the distance between x and one is less than some number, then the distance between f and 3 is less than some other "arbitrary number" it is just the math way of saying that one is close to the other

Answer 20

We should really have \[0<|x-1|<\delta\]

Answer 21

\[\left|f(x)-L \right|< \epsilon \Rightarrow \left|\frac{3x(x-1)}{x-1}-3 \right|<\epsilon\]
For x not equal to 1 we have:
\[\left| \frac{3x(x-1)}{x-1}-3 \right|<\epsilon \Rightarrow \left| 3x-3 \right|<\epsilon\]
\[\left| (3)(x-1) \right|<\epsilon \Rightarrow 3\left| x-1 \right|<\epsilon\Rightarrow \left| x-1 \right|<\frac{\epsilon}{3}\]
So by letting delta be:
\[\delta = \frac{\epsilon}{3}\]
you can guarantee the statement is true, and you are done.

Answer 22

What does the => sign mean?

Answer 23

"implies", i just use it to show my steps.

Answer 24

if you are confused with the epsilon,delta notation then try working out the problem with a specific epsilon like .1 and see what happens

Answer 25

But remember that epsilon is arbitrary and your proof must hold for any epsilon

Answer 26

Ok so Zarkon you are saying that I let delta be 0.1, and epsilon be whatever the corresponding y-value is for that, and test it out?

Answer 27

No hes saying try with a specific epsilon not a specific delta.

Answer 28

no...suppose I'm god and I am telling you that epsilon is .1

what is the range of values delta could be

Answer 29

Show that if epsilon is 0.1 you can find a delta

Answer 30

joemath314159 proof sorta gives it away ;)

Answer 31

im sry >.<

Answer 32

|x-c|<d -->|f(x)-L|<e <-- what I don't understand is how this makes sense. Also, what does the "--->" mean?

Answer 33

if 0<|x-c|<d then |f(x)-L|<e