Solving a limit with the ε-δ definition

[mHo2]

Homework Statement

I'm attempting to solve
Lim X^2 = 4
x-> 2
With the ε-δ definition of the limit. I have successfully solved simpler problems with the same model(first degree polynomials), but i cannot seem to get past some of the algebra/steps required to solve this limit, or by extension any higher-degree polynomial limit.

Homework Equations

0 < |x-c| < δ

|f(x) - L| < ε

Lim X^2 = 4
x-> 2


This is one example that i have been following off of, and i understand it completely up until
4minutes 10seconds, this step confuses me!

The Attempt at a Solution

I only looked for a video explaining this method after my book failed to help me out.
Here is the books solution(just the steps, not the side text) for solving this limit:

Lim x^2 = 4
x-> 2

Therefore:
0 < |x-2| < δ

|X^2 - 4| < ε
|x - 2||x + 2| < ε
X will be in the interval:
(1,3)
Therefore:
3 > x + 2 < 5
|x + 2| < 5 (why can i remove that lower bound if i add an absolute value? :S)

Let δ be the minimum of ε/5 and 1. (don't understand this part either!)

Then
|X^2 - 4| = |x - 2||x + 2| < (5)(ε/5) = ε

So it's mainly when the person solving the limit starts to mess around with the 'bounds' that i get messed up!

Thanks for any help.

- Mat

 

[Slats18]

Hi Mat =)

To point out a few things: the book didn't remove the lower bound because of the modulus being implement, we only need it's upper bound.

So by showing that |x+2| < 5, we can say that |x-2||x+2|< |x-2|*5 < ε <=> |x-2|< ε/5

Hence, take δ = min(ε/5 , 1). The reason 1 was chosen was because we needed our δ to be less than or equal to some number earlier (it's about the 2:30 mark on the youtube video). No fancy stuff involving the 1, it's just a chosen number haha.

And from there, complete the proof by showing that |x^2 - 4| < ε and you're done =)

That should be all correct, haven't done epsilon-delta proofs for bounded limits in a while haha, hope you can understand it =)

 

[mHo2]

Hi Mat =)

To point out a few things: the book didn't remove the lower bound because of the modulus being implement, we only need it's upper bound.

Okay, that makes sense, however my question is, why do we only want to deal with the upper bound and not the lower bound as well?

So by showing that |x+2| < 5, we can say that |x-2||x+2|< |x-2|*5 < ε <=> |x-2|< ε/5

I also think i understand this here, simply because if |x+2| has a maximum of 5 (rather infinitesimally close to 5) we can say that |x-2||x+2| will always be less than it..

Hence, take δ = min(ε/5 , 1). The reason 1 was chosen was because we needed our δ to be less than or equal to some number earlier (it's about the 2:30 mark on the youtube video). No fancy stuff involving the 1, it's just a chosen number haha.

I don't understand this part. Maybe it is because i haven't ever heard about this 'min' function before, can you expand on it?

Thanks!

- Mat

 

[SammyS]

Graph y=x². Pay particular attention to the graph from x=1 to x=3 .

 

[mHo2]

Graph y=x². Pay particular attention to the graph from x=1 to x=3 .

I understand what is going on at that point conceptually, but not algebraicly. So I'm not sure if graphing it will help me out..

 

[mHo2]

Bumping, any help?

Thanks!

 

[SammyS]

If you're not sure that graphing will help, then why not try it?

 

[mHo2]

If you're not sure that graphing will help, then why not try it?

I have tried graphing it, I was just being political.

Any help?

Bump!

 

[hunt_mat]

The value of 5 is taken because x is in the range of 2-delta and 2+delta, we can assume that delta<1, and therefore |x+2|<2+2+1