Question

prove only using the definition that lim as x approaches 0 of (x+1)^3 = 1. So I have "Let epsilon>0. Then choose some delta>0(need to find) such that if 0 12 years ago

Answer 1

Can you please help me?:

http://openstudy.com/updates/526082dae4b0c78c62eade37

Answer 2

I have notes somewhere on how to do this...

Answer 3

Give me a few minutes..

Answer 4

that would help so much, I think that it is just a lot of algebra manipulation and triangle identity but not sure how to manipulate it

Answer 5

You're right.. it's been a while since i did this.

Answer 6

Ok, here goes.. I'm gonna use e for epsilon and d for delta.

Answer 7

Given e>0, find d>0, such that when
0<|x-x0|<d,
|f(x) - L|<e

\[\lim_{x \rightarrow 0}(x+1)^3=1\]

Answer 8

Taking cube roots on both sides, we end up with:\[\lim_{x \rightarrow 0}(x+1)=1\]

Answer 9

So now we know that:\[x_0=-1, \space L=1\]

Answer 10

I do know that I have to start with l(x+1)^3-1l<e then manipulate to somehow get lxl<something*e

Answer 11

Yea.. im just finding this hard to follow while typing.. Hang on I'll attach my notes, look at page 4, i have like 3-4 examples

Answer 12

thanks for your time but I guess i just need to keep trying to rearrange it until I get what I need

Answer 13

Hey did you get it?

Answer 14

nope still working on it

Answer 15

Don't write all the symbols stick to the stuff that is important \[\begin{split}
|x-0|&<\delta &\implies |(x+1)^3-1|&<\epsilon \\
|x|&<\delta &\implies |(x^3+3x^2+3x+1)-1|&<\epsilon \\
|x|&<\delta &\implies |x^3+3x^2+3x|&<\epsilon \\
|x|&<\delta &\implies |x(x^2+3x+3)|&<\epsilon \\
|x|&<\delta &\implies |x||x^2+3x+3|&<\epsilon \\


\end{split}
\]

Answer 16

yep got that far then I don't know what to do

Answer 17

Pick a maximum delta range you want to allow.
Usually \(1\) is decent.
Then figure out what \(|x^2+3x+3|\) ranges for that value.

Answer 18

I think at this point you pick an arbitrary number for delta

Answer 19

lol wio got there before me..

Answer 20

so I would say let -1<x<1, then plug in -1 and 1 into the quadratic

Answer 21

Yeah, you want to plug in -1 and 1, and any other number you might think would be the maximum.

Answer 22

its probably 1 since we are approaching 0

Answer 23

It looks like the highest it will be is when \(x\) is \(1\).

Answer 24

It's an upwards pending parabola, so the further from it's vertex you get the higher it is.

Answer 25

does that really matter since we are looking at values arbitrarily close to 0

Answer 26

No, I'm giving you rigorous proof that \(x=1\) is the absolute maximum on the inverval \(-1<x<1\)

Answer 27

In which case \[
|(1)^2+3(1)+3| = 7
\]

Answer 28

oh ok. then we can say to choose d=min?(1,e/7)

Answer 29

Yes.

Answer 30

Well, we'd first say \[
7|x|<\epsilon\\
|x|<\frac \epsilon 7
\]

Answer 31

You want to be thorough on these, it will help you understand what you are doing better.

Answer 32

right, ok thank you so much. I am never too good when choosing mins and maxs

Answer 33

It is also very easy to make an error in logic and then "prove" something that is false.

Answer 34

So being thorough keeps your assumptions in check.
Anyway good luck.