Question

By using limit definition: prove x-->2 lim x^2 = 4
How to get epsilon for |x-2|.|x+2| < epsilon.
why would i put |x-2| < 1!!

Answer 1

\[
|x-2|\lt \delta \implies |x^2-4|<\epsilon\\
|x-2|\lt \delta \implies |x-2||x+2|<\epsilon\\
|x-2|\lt \delta \implies |x-2|<\frac{\epsilon}{|x+2|}\\
\]Ideally you would want to say \[
\delta = f(\epsilon) = \frac{\epsilon}{|x+2|}
\]However, you don't get to know what \(x\) will be, so you have to be a bit more clever.

Answer 2

What you can do in this case though, is to find some minimum delta range that you can always fallback on. In this case, we will let \(\delta=1\) be the fallback.

Answer 3

yea why 1

Answer 4

So we really only have to worry about cases where \(2-1\leq x \leq 2+1\).
This means that \(1\leq x \leq 3\).

Answer 5

To be honest, it could be any number grater than \(0\).

Answer 6

We could have chosen \(1/2000\). I chose \(1\) because I think it will be an easy number

Answer 7

so i can pick 5 or 10 my text always choose it to be 1

Answer 8

Yes, you can pick 5 or 10 if you want

Answer 9

Typically you want to pick a sufficiently small number.

Answer 10

Since \(1\) is easy to work with, a lot of times \(1\) will be picked.

Answer 11

However, do you mind if I actually just step back a bit and explain better what I am trying to accomplish?

Answer 12

nope ,but i know the rest i just wondered that text just picked one!!

Answer 13

Do you understand why they had to pick a number?

Answer 14

nope :D

Answer 15

Okay so let's go back to this for a moment:\[
\delta = f(\epsilon) = \frac{\epsilon}{|x+2|}
\]We don't really know what \(x\) is. We want to be safe. The safest we can possibly is to have \(\delta\) as small as possible. Thus what we want to say is: \[

\delta = f(\epsilon) = \min\left(\frac{\epsilon}{|x+2|}\right)

\]

Answer 16

It is always safer to make \(\delta\) smaller. The smaller it is, the smaller our \(\epsilon\) range will be.

Answer 17

Does this make sense?\[
\min\left(\frac{\epsilon}{|x+2|}\right) = \frac{\epsilon}{\max(|x+2|)}
\]

Answer 18

yea

Answer 19

The minimum is where the denominator is maximum.
The problem now is that there is no \(\max\) to \(|x+2|\). It could go to \(\infty\).

Answer 20

We can't pick \(\epsilon/\infty\) because \(\epsilon/\infty = 0 \) and we must pick \(\delta\) to be greater than \(0\).

Answer 21

So we have to pick a cutoff point.

Answer 22

In the book, they said \(\delta \le 1\). Since we get to chose what \(\delta\) is, we can always create a restriction like this.

Answer 23

The result of this is that \(|x-2|<1\implies 1\).
This means that \(\max(|x+2|) = 5\), and so: \[
\delta = \min\left(1,\frac{\epsilon }{5}\right)
\]

Would you like me to show what happens when we say \(\delta \le 5\)?

Answer 24

Does this is related with that the maximum is when x+2=1 ,so x =-1 ??

Answer 25

\[
|x-2|<5 \implies 2-5<x<2+5\implies -3<x<7\implies 0<|x+2|<9
\]

Answer 26

\[
\implies |x+2|<9
\]

Answer 27

-1 and 9

Answer 28

Well, the absolute values ensures it is greater that \(0\).

Answer 29

Or perhaps I should say greater than or equal

Answer 30

OK ,but does the one is related to the maximum is when x =-1

Answer 31

That max of what?

Answer 32

epsilon / x+2

Answer 33

It will max out when you approach \(-2\)

Answer 34

oh uh... I got wrong.
So that is write for prove to choose 1 ,but in real i would choose large number wright?

Answer 35

You want to chose \(\delta\) to be small, actually.

Answer 36

Unfortunately you can't chose \(\delta=0\).

Answer 37

Certainly, is would be safer to choose \(10^{-10000000}\) than choosing \(1=10^0\).

Answer 38

However, if you chose that, then you'd end up with: \[
|x-2|<10^{-1000000} \implies 2-10^{-1000000}<x<2+10^{-1000000}\\
\implies |x+2| < 4+10^{-1000000}
\]See...
It just makes it more complicated. It is easier to work with \(1\).

Answer 39

And since \(1\) is good enough, then might as well chose \(1\).
It really is that simple.

Answer 40

x-2 < min (e/x+2)
min when x ---> infinity and the left side grows also ????

Answer 41

what is wrong with that i am confused

Answer 42

Remember what epsilon really is....

Answer 43

\[
|f(x)-L|<e
\]

Answer 44

So you are saying: \[
|x-2| \leq \min\left(\frac{|x^2-4|}{|x+2|}\right)\lt \min\left(\frac{\epsilon}{|x+2|}\right)
\]

Answer 45

So yes, the left site will grow as well, because \(\epsilon\) range will have to grow.

Answer 46

As \(x\) departs from \(2\), \(f(x)\) departs from \(4\).

Answer 47

Well, potentially, anyway

Answer 48

OK . I think I got it now.
Thank you very much man .