Using min. when calculating limits

[brainy kevin]

I'm not really sure if this would go in homework or the calculus and analysis forums, so if I picked the wrong one please correct me. My calculus book has an example worked out that confused me, but you have to use the example to do the problems, which is why I put it here.

(Note: Since I don't know how to use LaTeX, I will just use the notation abs(x) for the absolute value of x, and e and o for epsilon and delta respectivly.)

Homework Statement

It is proving that the function f(x)=x² approaches 9 near 3. I'm going to start writing what the book says now.

We must ensure that abs(x²-9)<e for any given positive e. The obvious first step is to write: abs(x²-9)=abs(x-3)*abs(x+3). We require that abs(x-3)<1, which gives us abs(x+3)<7, which gives us:

abs(x-3)*abs(x+3)<7abs(x-3) [I verified that this inequality is true, everything up to this point makes perfect sense to me, we're just choosing our specific deltas.]

This shows us we have abs(x²-9)<e for abs(x-3)<1, [Yes it does] and abs(x-3)<e/7; or officially, we require abs(x-3)<min(e/7,1).

Wait just a second.

Homework Equations

Since this is a proof, there really aren't any.

The Attempt at a Solution

I understand everything before my comment, "yes it does." However, if e=7abs(x-3), then e/7 would be abs(x-3), which is saying abs(x-3)<abs(x-3), which is ridiculous! Is this a typo, or did I truly miss something? Thank you for your help.

 

[Mark44]

The idea is that, if someone gives you an \epsilon, your job is to find a \delta &gt; 0 so that if x is within \delta of 3, then x² will be within \epsilon of 9.

The trouble is, you don't know in advance whether this person will give you a small number for \epsilon, and this can make a difference in what you specify for \delta. For this reason, you need to choose the smaller of {1, \epsilon/7} as your \delta.

So try it out with these values of \epsilon that I specify:
1) \epsilon = 14. What should \delta be?
2) \epsilon = 0.1. What should \delta be?

For each of these examples, show that if |x - 3| < \delta, then |x² - 9| < \epsilon.

 

[brainy kevin]

I think you misunderstood my question. I'm asking why it looks like it's saying that the absolute value of x minus three is less than the absolute value of x minus three, which is a clear contradiction. I understand the theory behind limits very well, it's the part where you use the minimum function that I don't understand. I appreciate your help though.

 

[snipez90]

or officially, we require abs(x-3)<min(e/7,1).

Yes, we do. This means that abs(x-3) is less than the smaller of e/7 and 1.

However, if e=7abs(x-3), then e/7 would be abs(x-3), which is saying abs(x-3)<abs(x-3), which is ridiculous! Is this a typo, or did I truly miss something? Thank you for your help.

Well the whole idea is that e is an arbitrarily small positive number that doesn't depend on x. This ensures that we do not get the contradiction you described. The definition of the limit is really a static one. We don't let epsilon depend on another variable, but rather let it be a number that serves to prescribe an open interval (a band of width 2e) that can be made arbitrarily small. Sorry I misread your initial question.

*EDIT* Note that here, it is possible to find the largest value of delta by rewriting |x+3| as |x - 3 + 6| and applying the triangle inequality. This will allow you to find delta in terms of epsilon without the need for the min function. However, applying the min function is a lot more flexible. Although the min function has a formula, we are more interested in knowing that it ensures that multiple (often strict) inequalities are held. In an ordered set, if we need a < b and a < c, we might as well require a < min(b,c).