# Homework Help: Limits - Apostol book

1. Mar 18, 2013

### mamute

Hello guys, I am stuck in page 129 on Calculus Vol I - Apostol book. I would like to know if there is anybody here who can help me. I am not a mathematician, so It might be a simple transformation but I am not going through it.

He states that:

lim(x->p) f(x) = A is equivalent to say that:

lim(x->p) ( f(x) - A) = 0 OR

lim(x->p) | f(x) - A| = 0

I can not make these transformations algebraically, how can it be done?

Thanks for any help
Cheers

2. Mar 18, 2013

### Dick

Would you agree a=b, a-b=0 and |a-b|=0 all mean the same thing? That's really almost the whole story. There is nothing complicated going on here.

Last edited: Mar 18, 2013
3. Mar 18, 2013

### mamute

The book's definition of a limit is such that:

| f(x) -A | < ε whenever 0 < | x - p | < δ

So it presents the three equations (on the previous post) and it says: The equivalence becomes apparent as soon as we write each of these statements in the ε, δ terminology.

That is the core of my doubts. How to write those three equations.

Sorry that I wasn't clear enough in the first post.

4. Mar 18, 2013

### Dick

The definition of lim(x->p) f(x)=A is for all ε>0 there exists a δ>0 such that |f(x)-A|<ε whenever 0<|x-p|< δ. Don't leave out the quantifiers. Now follow their advice and write out what lim(x->p) (f(x)-A)=0 means.

5. Mar 18, 2013

### mamute

that is exactly the point where I am confused.

Since we have:
lim(x->p) f(x) = A, then, in my opinion, we would have

lim(x->p) ( f(x) - A) = 0, since A is defined in the line above:

lim(x->p) ( f(x) - lim(x->p) f(x) ) = 0

And I don't know how to go any further from here. ;/

Alternatively, iff you consider:
lim(x->p) ( f(x) -A ) = 0, as we know |f(x) -A| < ε
lim(x->p) ( ε ) would be ε, not zero.

Last edited: Mar 19, 2013
6. Mar 19, 2013

### Dick

If lim(x->p) g(x)=0 then what's the definition of that? Spell it out for me. Then put g(x)=f(x)-A.

7. Mar 19, 2013

### mamute

Thank you, that was good.